💡First, I need to determine the distance the man covers while crossing the bridge. He walks at a speed of 10 kilometers per hour and takes 15 minutes to cross.To make the units consistent, I'll convert the time from minutes to hours. There are 60 minutes in an hour, so 15 minutes is 15/60 hours, which simplifies to 0.25 hours.Next, I'll use the formula for distance, which is speed multiplied by time. Plugging in the values, the distance is 10 km/h multiplied by 0.25 hours, resulting in 2.5 kilometers.Finally, to express the length of the bridge in meters, I'll convert kilometers to meters. Since 1 kilometer equals 1000 meters, 2.5 kilometers is equal to 2500 meters.

💡First, I need to determine how much Rebecca charges for a dye job. I'll start by calculating her total earnings from haircuts and perms.She has four haircuts at 30 each, which totals 120. She also has one perm at 40, adding another 40 to her earnings. This gives a combined total of 160 from haircuts and perms.Next, I'll consider the tips she received, which are 50. Adding this to the 160 from services, the total becomes 210.Rebecca's total earnings for the day are 310. To find out how much she made from dye jobs, I'll subtract the earnings from haircuts, perms, and tips from the total earnings: 310 - 210 = 100.This 100 represents the profit from the two dye jobs after accounting for the cost of hair dye. Since each dye job requires a 10 box of hair dye, the total cost for two dye jobs is 20.To find the total amount charged for the two dye jobs before costs, I'll add the profit and the cost of dye: 100 + 20 = 120.Finally, to determine the charge per dye job, I'll divide the total amount by the number of dye jobs: 120 / 2 = 60.Therefore, Rebecca charges 60 for a dye job.

💡Alright, so I have this trigonometry problem here. It says that cos(π/6 - α) equals √3/3, and I need to find the value of cos(5π/6 + α) plus sin²(α - π/6). Hmm, okay, let's break this down step by step.First, I remember that there are some trigonometric identities that might help here. The problem involves angles that are related, so maybe I can use some angle addition or subtraction formulas. Let me recall those. The cosine of a sum formula is cos(A + B) = cosA cosB - sinA sinB, and the sine of a difference formula is sin(A - B) = sinA cosB - cosA sinB. Maybe these will come in handy.Looking at the first part, cos(5π/6 + α). I wonder if I can express this in terms of the given angle, which is π/6 - α. Let me see. 5π/6 is equal to π - π/6. So, 5π/6 + α is the same as π - π/6 + α. That simplifies to π + (α - π/6). Hmm, interesting. So, cos(5π/6 + α) is the same as cos(π + (α - π/6)).I remember that cos(π + θ) is equal to -cosθ. So, applying that identity here, cos(π + (α - π/6)) becomes -cos(α - π/6). Wait, but in the problem, we have cos(π/6 - α). Is that the same as cos(α - π/6)? Let me think. Cosine is an even function, so cos(-θ) = cosθ. Therefore, cos(π/6 - α) is the same as cos(α - π/6). So, cos(5π/6 + α) is equal to -cos(π/6 - α). Given that cos(π/6 - α) is √3/3, then cos(5π/6 + α) is -√3/3. Okay, that takes care of the first part.Now, moving on to the second part: sin²(α - π/6). I need to find the square of the sine of (α - π/6). Again, since sine is an odd function, sin(-θ) = -sinθ, so sin(α - π/6) is equal to -sin(π/6 - α). Therefore, sin²(α - π/6) is equal to [sin(π/6 - α)]², which is the same as sin²(π/6 - α).I know that sin²θ can be expressed using the identity sin²θ = 1 - cos²θ. So, sin²(π/6 - α) is equal to 1 - cos²(π/6 - α). We already know that cos(π/6 - α) is √3/3, so cos²(π/6 - α) is (√3/3)², which is 3/9, simplifying to 1/3.Therefore, sin²(π/6 - α) is 1 - 1/3, which equals 2/3. So, sin²(α - π/6) is also 2/3.Now, putting it all together: cos(5π/6 + α) + sin²(α - π/6) is equal to -√3/3 + 2/3. To combine these, I can write them over a common denominator. Both terms have a denominator of 3, so it becomes (-√3 + 2)/3.Wait, but the problem asks for the expression in a box, so I think that's the final answer. Let me just double-check my steps to make sure I didn't make any mistakes.1. I recognized that 5π/6 is π - π/6, so I rewrote cos(5π/6 + α) as cos(π + (α - π/6)).2. Using the identity cos(π + θ) = -cosθ, I found that this is equal to -cos(α - π/6).3. Since cosine is even, cos(α - π/6) is the same as cos(π/6 - α), which is given as √3/3. So, cos(5π/6 + α) is -√3/3.4. For sin²(α - π/6), I used the identity sin²θ = 1 - cos²θ. Since sin²(α - π/6) is the same as sin²(π/6 - α), I substituted cos(π/6 - α) = √3/3 and found that sin²(π/6 - α) is 2/3.5. Adding the two results together: -√3/3 + 2/3, which simplifies to (2 - √3)/3.Wait a second, I think I made a mistake in the last step. When I combined -√3/3 and 2/3, I should have written it as (2 - √3)/3, not (2 - √3)/3. Hmm, actually, that's correct. Both terms have a denominator of 3, so combining them gives (2 - √3)/3.But looking back at the problem, the answer needs to be in a box. So, I think my final answer is (2 - √3)/3. Let me just make sure I didn't make any calculation errors.Starting from the beginning:Given cos(π/6 - α) = √3/3.We need to find cos(5π/6 + α) + sin²(α - π/6).1. cos(5π/6 + α) = cos(π - π/6 + α) = cos(π + (α - π/6)) = -cos(α - π/6) = -cos(π/6 - α) = -√3/3.2. sin²(α - π/6) = sin²(π/6 - α) = 1 - cos²(π/6 - α) = 1 - (√3/3)² = 1 - 1/3 = 2/3.Adding them together: -√3/3 + 2/3 = (2 - √3)/3.Yes, that seems correct. So, the final answer is (2 - √3)/3, which can be written as boxed{dfrac{2 - sqrt{3}}{3}}.

💡Okay, so I have this problem here: I need to find the imaginary part of the complex number 4i divided by (1 - i). The options are A: -2, B: 2, C: 2i, and D: -2i. Hmm, let me think about how to approach this.First, I remember that when dealing with complex numbers in the denominator, it's often helpful to multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of (1 - i) is (1 + i). This should help simplify the expression.So, let me write that down:[frac{4i}{1 - i} times frac{1 + i}{1 + i}]Multiplying the numerators together and the denominators together:Numerator: 4i * (1 + i) = 4i + 4i^2Denominator: (1 - i)(1 + i) = 1^2 - (i)^2 = 1 - (-1) = 2Wait, so the denominator simplifies to 2. That's good. Now, let's work on the numerator:4i + 4i^2I know that i^2 is equal to -1, so substituting that in:4i + 4(-1) = 4i - 4So now, the entire expression becomes:[frac{4i - 4}{2}]I can separate this into real and imaginary parts:[frac{-4}{2} + frac{4i}{2} = -2 + 2i]So, the complex number simplifies to -2 + 2i. Now, the question asks for the imaginary part. The imaginary part is the coefficient of i, which in this case is 2.Let me double-check my steps to make sure I didn't make a mistake. I multiplied numerator and denominator by the conjugate, expanded both, simplified using i^2 = -1, and then separated the real and imaginary parts. It all seems to add up.Wait, just to be thorough, let me verify the multiplication step:(1 - i)(1 + i) = 1*1 + 1*i - i*1 - i*i = 1 + i - i - i^2 = 1 - (-1) = 2. Yep, that's correct.And the numerator: 4i*(1 + i) = 4i + 4i^2 = 4i - 4. That looks right too.So, after simplifying, the expression is -2 + 2i, and the imaginary part is indeed 2. So the answer should be B: 2.I think that's solid. I don't see any mistakes in my calculations, so I feel confident about this answer.

💡Okay, so I have this problem where I need to find the number of functions ( f : mathbb{R} to mathbb{R} ) such that[ f(x + f(y)^2) = x + y^2 ]for all real numbers ( x ) and ( y ). Hmm, functional equations can be tricky, but let me try to break this down step by step.First, I remember that when dealing with functional equations, a common strategy is to plug in specific values for ( x ) or ( y ) to simplify the equation and get more information about ( f ). Let me try setting ( x = 0 ) because that often simplifies things.So, if I set ( x = 0 ), the equation becomes:[ f(0 + f(y)^2) = 0 + y^2 ]which simplifies to:[ f(f(y)^2) = y^2 ]Alright, that's interesting. This tells me that applying ( f ) twice on ( f(y)^2 ) gives me ( y^2 ). So, ( f ) composed with ( f(y)^2 ) is just squaring ( y ). Maybe this suggests that ( f ) is related to squaring or square roots? Hmm, not sure yet.Let me think about possible functions ( f ) that could satisfy this. The simplest functions to test are linear functions, like ( f(y) = y ) or ( f(y) = -y ). Let me try these.First, let's assume ( f(y) = y ). Then, substituting back into the original equation:[ f(x + f(y)^2) = f(x + y^2) = x + y^2 ]And the right side is:[ x + y^2 ]So, yes, it works! That means ( f(y) = y ) is indeed a solution.Now, let's try ( f(y) = -y ). Substituting into the original equation:[ f(x + f(y)^2) = f(x + (-y)^2) = f(x + y^2) ]Since ( (-y)^2 = y^2 ), this simplifies to:[ f(x + y^2) = x + y^2 ]But wait, ( f(x + y^2) = -(x + y^2) ) if ( f(y) = -y ). So, substituting back, we get:[ -(x + y^2) = x + y^2 ]Which simplifies to:[ -x - y^2 = x + y^2 ]Bringing all terms to one side:[ -x - y^2 - x - y^2 = 0 ][ -2x - 2y^2 = 0 ]Which implies:[ -2x - 2y^2 = 0 ]Dividing both sides by -2:[ x + y^2 = 0 ]But this has to hold for all real numbers ( x ) and ( y ), which is impossible because ( x ) and ( y ) can be any real numbers. So, this suggests that ( f(y) = -y ) doesn't satisfy the original equation. Wait, that's confusing because earlier when I set ( x = 0 ), it seemed like ( f(f(y)^2) = y^2 ) worked for ( f(y) = -y ). Maybe I made a mistake in my substitution.Let me double-check. If ( f(y) = -y ), then ( f(y)^2 = (-y)^2 = y^2 ). So, ( f(f(y)^2) = f(y^2) = -y^2 ). But according to the equation when ( x = 0 ), ( f(f(y)^2) = y^2 ). So, ( -y^2 = y^2 ), which implies ( y^2 = 0 ) for all ( y ), which is only true if ( y = 0 ). But the equation must hold for all ( y ), so this is a contradiction. Therefore, ( f(y) = -y ) is not a solution. Hmm, so my initial thought that ( f(y) = -y ) works was incorrect.Wait, maybe I confused something. Let me try substituting ( f(y) = -y ) again into the original equation:[ f(x + f(y)^2) = f(x + (-y)^2) = f(x + y^2) ]Since ( f(z) = -z ), this becomes:[ -(x + y^2) ]But the right side of the original equation is:[ x + y^2 ]So, we have:[ -(x + y^2) = x + y^2 ]Which simplifies to:[ -x - y^2 = x + y^2 ]Bringing like terms together:[ -x - x = y^2 + y^2 ][ -2x = 2y^2 ][ -x = y^2 ]Again, this must hold for all ( x ) and ( y ), which is impossible because ( y^2 ) is always non-negative, but ( -x ) can be any real number depending on ( x ). So, this is a contradiction, meaning ( f(y) = -y ) is not a solution.Okay, so scratch that. So far, only ( f(y) = y ) seems to work. Maybe there are other functions besides the identity function? Let me explore further.Looking back at the equation ( f(f(y)^2) = y^2 ), this suggests that ( f ) is surjective on the non-negative real numbers because for any ( y^2 geq 0 ), there exists some ( z = f(y)^2 ) such that ( f(z) = y^2 ). So, ( f ) maps some real number ( z ) to ( y^2 ), which is non-negative. Therefore, the image of ( f ) must cover all non-negative real numbers.Also, from the original equation, if I fix ( y ) and vary ( x ), the left side is ( f(x + f(y)^2) ), and the right side is ( x + y^2 ). This suggests that ( f ) is linear in some way because as ( x ) increases, the output increases linearly. Maybe ( f ) is affine or linear.Let me assume that ( f ) is linear, so let me suppose ( f(y) = ky + c ) for some constants ( k ) and ( c ). Let's see if this works.Substituting into the original equation:[ f(x + f(y)^2) = k(x + f(y)^2) + c = kx + k(f(y)^2) + c ]And the right side is:[ x + y^2 ]So, equating the two:[ kx + k(f(y)^2) + c = x + y^2 ]This must hold for all ( x ) and ( y ). So, let's equate coefficients.First, the coefficients of ( x ):[ k = 1 ]So, ( k = 1 ).Now, the constant term:[ k(f(y)^2) + c = y^2 ]But ( k = 1 ), so:[ f(y)^2 + c = y^2 ]But ( f(y) = ky + c = y + c ), so:[ (y + c)^2 + c = y^2 ]Expanding:[ y^2 + 2cy + c^2 + c = y^2 ]Subtracting ( y^2 ) from both sides:[ 2cy + c^2 + c = 0 ]This must hold for all ( y ), which implies that the coefficients of ( y ) and the constant term must both be zero.So, coefficient of ( y ):[ 2c = 0 implies c = 0 ]Constant term:[ c^2 + c = 0 ]But since ( c = 0 ), this gives:[ 0 + 0 = 0 ]Which is true.Therefore, the only linear solution is ( f(y) = y ). So, that's consistent with our earlier result.But wait, could there be non-linear solutions? The problem doesn't specify that ( f ) has to be linear, so I shouldn't limit myself to that.Let me think about the equation again:[ f(x + f(y)^2) = x + y^2 ]If I fix ( y ) and let ( x ) vary, the left side is ( f ) evaluated at ( x + f(y)^2 ), and the right side is a linear function in ( x ). This suggests that ( f ) is linear when restricted to arguments of the form ( x + f(y)^2 ). But does this imply that ( f ) is linear everywhere?Alternatively, maybe ( f ) is additive or something. Let me explore that.Suppose ( f(a + b) = f(a) + f(b) ) for all ( a, b ). Then, ( f ) is additive. But I don't know if that's necessarily the case here.Wait, let me consider another substitution. What if I set ( y = 0 )? Let's try that.Setting ( y = 0 ):[ f(x + f(0)^2) = x + 0^2 ]Simplifies to:[ f(x + f(0)^2) = x ]Hmm, so ( f ) evaluated at ( x + c ) (where ( c = f(0)^2 )) is equal to ( x ). That suggests that ( f ) is a linear function with slope 1 when shifted by ( c ).Let me denote ( c = f(0)^2 ). Then, the equation becomes:[ f(x + c) = x ]So, if I let ( z = x + c ), then ( x = z - c ), and the equation becomes:[ f(z) = z - c ]So, ( f(z) = z - c ) for all ( z ). Therefore, ( f ) is a linear function with slope 1 and intercept ( -c ).But earlier, when I assumed ( f ) was linear, I found that ( c = 0 ), so ( f(z) = z ). So, this suggests that ( c = 0 ), meaning ( f(0)^2 = 0 ), so ( f(0) = 0 ).Therefore, ( f(z) = z ) is the only solution. So, that's consistent.Wait, but let me make sure. If ( f(z) = z - c ), then ( f(0) = -c ). But earlier, we have ( f(0)^2 = c ). So, substituting ( f(0) = -c ) into ( f(0)^2 = c ), we get:[ (-c)^2 = c ][ c^2 = c ][ c^2 - c = 0 ][ c(c - 1) = 0 ]So, ( c = 0 ) or ( c = 1 ).If ( c = 0 ), then ( f(z) = z ), which we already know works.If ( c = 1 ), then ( f(z) = z - 1 ). Let's test this function in the original equation to see if it works.So, ( f(z) = z - 1 ). Let's substitute into the original equation:[ f(x + f(y)^2) = f(x + (y - 1)^2) = (x + (y - 1)^2) - 1 = x + (y^2 - 2y + 1) - 1 = x + y^2 - 2y ]But the right side of the original equation is:[ x + y^2 ]So, we have:[ x + y^2 - 2y = x + y^2 ]Subtracting ( x + y^2 ) from both sides:[ -2y = 0 ]Which implies ( y = 0 ). But this must hold for all ( y ), which is a contradiction. Therefore, ( c = 1 ) does not lead to a valid solution.Thus, the only possible value is ( c = 0 ), which gives ( f(z) = z ). So, ( f(z) = z ) is the only solution.Wait, but earlier when I thought ( f(y) = -y ) might work, I realized it doesn't. So, is ( f(y) = y ) the only solution?Let me see if there are any other possibilities. Maybe ( f ) is not linear but still satisfies the equation.Suppose ( f ) is a quadratic function, say ( f(y) = ay^2 + by + c ). Let's see if that could work.But before I dive into that, let me think about the equation ( f(f(y)^2) = y^2 ). If ( f ) is quadratic, then ( f(f(y)^2) ) would be a quartic function, but the right side is quadratic. So, unless the higher degree terms cancel out, which seems unlikely, this probably doesn't work. So, maybe ( f ) can't be quadratic.Alternatively, maybe ( f ) is a square root function? But ( f ) maps from ( mathbb{R} ) to ( mathbb{R} ), and square roots are only defined for non-negative numbers, but the problem states ( f : mathbb{R} to mathbb{R} ), so ( f ) must be defined for all real numbers, including negatives. So, square root functions are out.What about absolute value? Let me test ( f(y) = |y| ).Substituting into the original equation:[ f(x + f(y)^2) = f(x + |y|^2) = |x + y^2| ]But the right side is:[ x + y^2 ]So, we have:[ |x + y^2| = x + y^2 ]Which is true only if ( x + y^2 geq 0 ). But since ( x ) can be any real number, including negative numbers, this isn't always true. For example, if ( x = -1 ) and ( y = 0 ), then ( | -1 + 0 | = 1 ), but the right side is ( -1 + 0 = -1 ), which is not equal. So, ( f(y) = |y| ) doesn't work.Hmm, so maybe ( f ) has to be linear. Since we've already found that ( f(y) = y ) is the only linear solution, and non-linear functions seem problematic, perhaps that's the only solution.Wait, let me think differently. From the equation ( f(x + f(y)^2) = x + y^2 ), if I fix ( y ) and let ( x ) vary, then for each fixed ( y ), the function ( f ) is linear in ( x ). Specifically, ( f ) must satisfy:[ f(x + c) = x + d ]where ( c = f(y)^2 ) and ( d = y^2 ). So, for each ( y ), ( f ) is linear when shifted by ( c ). But this has to hold for all ( y ), which suggests that ( f ) is linear everywhere.Therefore, ( f ) must be a linear function. As we saw earlier, the only linear function that satisfies the equation is ( f(y) = y ).Wait, but let me make sure. Suppose ( f ) is linear, so ( f(y) = ky + c ). Earlier, we saw that ( k = 1 ) and ( c = 0 ). So, ( f(y) = y ) is the only linear solution.But could there be a non-linear solution? Let me suppose that ( f ) is non-linear but still satisfies the equation.Suppose ( f(y) = y ) for ( y geq 0 ) and ( f(y) = -y ) for ( y < 0 ). Let's test this.So, for ( y geq 0 ), ( f(y) = y ), and for ( y < 0 ), ( f(y) = -y ). Let's substitute into the original equation.Case 1: ( y geq 0 ). Then, ( f(y) = y ), so:[ f(x + y^2) = x + y^2 ]Which holds because ( f(z) = z ) for ( z geq 0 ), and ( x + y^2 ) can be any real number depending on ( x ). Wait, but ( x ) can be negative, so ( x + y^2 ) could be negative. If ( x + y^2 < 0 ), then ( f(x + y^2) = -(x + y^2) ), but the right side is ( x + y^2 ). So, we have:[ -(x + y^2) = x + y^2 ]Which implies ( -x - y^2 = x + y^2 ), leading to ( -2x - 2y^2 = 0 ), or ( x + y^2 = 0 ). But this must hold for all ( x ) and ( y ), which is impossible. Therefore, this piecewise function doesn't work.Case 2: ( y < 0 ). Then, ( f(y) = -y ), so:[ f(x + (-y)^2) = f(x + y^2) ]Which is the same as:[ f(x + y^2) = x + y^2 ]But again, if ( x + y^2 < 0 ), ( f(x + y^2) = -(x + y^2) ), leading to the same contradiction as above.Therefore, this piecewise function doesn't satisfy the equation for all ( x ) and ( y ).Hmm, so maybe any non-linear function would run into similar issues where the function's behavior on negative numbers would cause contradictions. Therefore, perhaps ( f(y) = y ) is indeed the only solution.Wait, let me think about another approach. From the equation ( f(f(y)^2) = y^2 ), we can see that ( f ) must be surjective on the non-negative real numbers because for any ( y^2 geq 0 ), there exists some ( z = f(y)^2 ) such that ( f(z) = y^2 ). So, ( f ) must map some real number ( z ) to every non-negative real number.Additionally, from the original equation, if I fix ( y ) and vary ( x ), the left side is ( f(x + f(y)^2) ), and the right side is ( x + y^2 ). This suggests that ( f ) is linear in ( x ) when shifted by ( f(y)^2 ). So, for each fixed ( y ), ( f ) is linear in ( x ). But since this must hold for all ( y ), ( f ) must be linear everywhere.Therefore, combining these observations, ( f ) must be a linear function, and the only linear function that satisfies the equation is ( f(y) = y ).Wait, but earlier when I considered ( f(y) = -y ), it didn't work because it led to a contradiction. So, is ( f(y) = y ) the only solution?Let me try another approach. Suppose ( f ) is bijective. From ( f(f(y)^2) = y^2 ), if ( f ) is bijective, then ( f(y)^2 ) must be unique for each ( y ). But since ( f(y)^2 ) is always non-negative, and ( f ) is bijective, ( f ) must map both positive and negative numbers to cover all real numbers. But ( f(y)^2 ) is non-negative, so ( f ) must map some real numbers to non-negative numbers and others to negative numbers, but ( f(f(y)^2) = y^2 ) is always non-negative. Hmm, not sure if this helps.Alternatively, let me consider the function ( f ) being injective or surjective.From ( f(f(y)^2) = y^2 ), since ( y^2 ) can take any non-negative value, ( f ) must be surjective onto the non-negative real numbers. So, ( f ) is surjective on ( [0, infty) ).Also, if ( f ) is injective, then ( f(a) = f(b) ) implies ( a = b ). Let's see if ( f ) is injective.Suppose ( f(a) = f(b) ). Then, for any ( x ), we have:[ f(x + f(a)^2) = x + a^2 ]and[ f(x + f(b)^2) = x + b^2 ]But since ( f(a) = f(b) ), ( f(x + f(a)^2) = f(x + f(b)^2) ), so:[ x + a^2 = x + b^2 ]Which implies ( a^2 = b^2 ), so ( a = b ) or ( a = -b ). Therefore, ( f ) is injective only if ( a = b ), but if ( a = -b ), then ( f(a) = f(b) ) would imply ( a = -b ), but ( f(a) = f(b) ) doesn't necessarily lead to ( a = b ). So, ( f ) is not necessarily injective unless ( a = b ).Wait, maybe I'm complicating things. Let me go back to the original equation and see if I can express ( f ) in terms of itself.From ( f(x + f(y)^2) = x + y^2 ), if I let ( z = x + f(y)^2 ), then ( x = z - f(y)^2 ), and substituting back, we get:[ f(z) = z - f(y)^2 + y^2 ]But this must hold for all ( z ) and ( y ). So, for a fixed ( z ), ( f(z) ) depends on ( y ), which is arbitrary. The only way this can hold is if the term involving ( y ) is constant, i.e., ( -f(y)^2 + y^2 ) is the same for all ( y ).Let me denote ( c = -f(y)^2 + y^2 ). Then, ( f(z) = z + c ) for all ( z ). So, ( f ) is a linear function with slope 1 and intercept ( c ).But earlier, we saw that if ( f(z) = z + c ), then substituting back into the equation ( f(f(y)^2) = y^2 ), we get:[ f(f(y)^2) = f(y^2 + c) = y^2 + c + c = y^2 + 2c ]But this must equal ( y^2 ), so:[ y^2 + 2c = y^2 ]Which implies ( 2c = 0 ), so ( c = 0 ). Therefore, ( f(z) = z ) is the only solution.Wait, this seems to confirm that ( f(z) = z ) is the only solution. So, putting it all together, the only function that satisfies the given equation is ( f(y) = y ).Just to make sure, let me test ( f(y) = y ) in the original equation one more time:[ f(x + f(y)^2) = f(x + y^2) = x + y^2 ]Which matches the right side exactly. So, yes, it works.Therefore, after considering various possibilities and approaches, it seems that the only function ( f ) that satisfies the equation is the identity function ( f(y) = y ).

💡Okay, so I have this function ( g(x) = 2^{x^2 - 4} - 2|x + 1| ), and I need to figure out if it's even, odd, or neither. Hmm, let me remember what even and odd functions are. An even function satisfies ( g(-x) = g(x) ) for all ( x ) in its domain. That means if I plug in ( -x ) into the function, it should give me the same result as plugging in ( x ). On the other hand, an odd function satisfies ( g(-x) = -g(x) ). So, if I plug in ( -x ), I should get the negative of the original function. If neither of these conditions hold, then the function is neither even nor odd.Alright, so let me try to compute ( g(-x) ) and see what happens. Starting with the function:[ g(x) = 2^{x^2 - 4} - 2|x + 1| ]So, substituting ( -x ) in place of ( x ), we get:[ g(-x) = 2^{(-x)^2 - 4} - 2|-x + 1| ]Wait, hold on, is that right? Let me double-check. The first term is ( 2^{(-x)^2 - 4} ). Since ( (-x)^2 ) is the same as ( x^2 ), because squaring a negative number makes it positive. So, that term simplifies to ( 2^{x^2 - 4} ). Now, the second term is ( -2|-x + 1| ). Hmm, let me think about the absolute value part. The expression inside the absolute value is ( -x + 1 ), which can also be written as ( -(x - 1) ). But absolute value of a number is the same as the absolute value of its negative, so ( |-x + 1| = |x - 1| ). Wait, but in the original function, the absolute value term is ( |x + 1| ). So, in ( g(-x) ), it's ( | -x + 1 | ), which is ( |1 - x| ). Is ( |1 - x| ) the same as ( |x + 1| )? Let me test with some numbers. If ( x = 2 ), then ( |1 - 2| = 1 ) and ( |2 + 1| = 3 ). They are not the same. So, that means ( |1 - x| ) is not equal to ( |x + 1| ) in general. Wait, so does that mean the second term in ( g(-x) ) isn't the same as in ( g(x) )? Let me write it out again:[ g(-x) = 2^{x^2 - 4} - 2| -x + 1 | ][ = 2^{x^2 - 4} - 2|1 - x| ]But in the original function, it's ( -2|x + 1| ). So, unless ( |1 - x| = |x + 1| ), which they aren't, as I saw with ( x = 2 ), then ( g(-x) ) isn't equal to ( g(x) ). Wait, but maybe I made a mistake in the substitution. Let me go back. The original function is ( -2|x + 1| ). So when I substitute ( -x ), it becomes ( -2| -x + 1 | ). Which is ( -2|1 - x| ). So, ( g(-x) = 2^{x^2 - 4} - 2|1 - x| ). Comparing this to ( g(x) = 2^{x^2 - 4} - 2|x + 1| ), the first terms are the same, but the second terms are different because ( |1 - x| ) is not the same as ( |x + 1| ). Therefore, ( g(-x) ) is not equal to ( g(x) ), so the function isn't even. Now, let's check if it's odd. For a function to be odd, ( g(-x) ) should equal ( -g(x) ). So, let's compute ( -g(x) ):[ -g(x) = -left(2^{x^2 - 4} - 2|x + 1|right) ][ = -2^{x^2 - 4} + 2|x + 1| ]Compare this to ( g(-x) ):[ g(-x) = 2^{x^2 - 4} - 2|1 - x| ]So, for ( g(-x) ) to be equal to ( -g(x) ), the following must hold:[ 2^{x^2 - 4} - 2|1 - x| = -2^{x^2 - 4} + 2|x + 1| ]Let me rearrange this equation:[ 2^{x^2 - 4} + 2^{x^2 - 4} = 2|x + 1| + 2|1 - x| ][ 2 cdot 2^{x^2 - 4} = 2(|x + 1| + |1 - x|) ][ 2^{x^2 - 3} = |x + 1| + |1 - x| ]Hmm, is this true for all ( x )? Let me test with some values.Take ( x = 0 ):Left side: ( 2^{0 - 3} = 2^{-3} = 1/8 )Right side: ( |0 + 1| + |1 - 0| = 1 + 1 = 2 )Not equal. So, the equation doesn't hold for ( x = 0 ). Therefore, ( g(-x) neq -g(x) ), so the function isn't odd either.Wait, but maybe I made a mistake in the substitution or the logic. Let me double-check.Starting again, ( g(-x) = 2^{x^2 - 4} - 2|1 - x| ). And ( -g(x) = -2^{x^2 - 4} + 2|x + 1| ). So, unless ( 2^{x^2 - 4} = -2^{x^2 - 4} ), which is only possible if ( 2^{x^2 - 4} = 0 ), which never happens, these two expressions can't be equal. So, ( g(-x) ) is not equal to ( -g(x) ).Therefore, the function is neither even nor odd.Wait, but earlier, I thought the first term was the same, but the second term was different. So, perhaps the function is neither. But let me think again.Wait, maybe I made a mistake in simplifying ( | -x + 1 | ). Let me see. ( | -x + 1 | = |1 - x| ), which is the same as ( |x - 1| ). So, ( |1 - x| = |x - 1| ). So, in the function ( g(-x) ), the second term is ( -2|x - 1| ). In the original function, it's ( -2|x + 1| ). So, unless ( |x - 1| = |x + 1| ), which is only true if ( x = 0 ), but not for all ( x ), then ( g(-x) ) is not equal to ( g(x) ). Therefore, the function isn't even. And since ( g(-x) ) isn't equal to ( -g(x) ), it's also not odd. So, the function is neither even nor odd.Wait, but in the initial problem, the assistant said it was even. Did I make a mistake? Let me check the assistant's solution.The assistant wrote:[ g(-x) = 2^{(-x)^2 - 4} - 2|-x - 1| ][ = 2^{x^2 - 4} - 2|x + 1| ][ = g(x) ]Wait, hold on, the assistant wrote ( |-x - 1| ), but in reality, substituting ( -x ) into ( |x + 1| ) gives ( |-x + 1| ), not ( |-x - 1| ). So, the assistant made a mistake in substitution. Because ( g(x) = 2^{x^2 - 4} - 2|x + 1| ), so ( g(-x) = 2^{(-x)^2 - 4} - 2| -x + 1 | ), which is ( 2^{x^2 - 4} - 2|1 - x| ), not ( 2^{x^2 - 4} - 2|x + 1| ). So, the assistant incorrectly wrote ( |-x - 1| ) instead of ( |-x + 1| ). Therefore, their conclusion that ( g(-x) = g(x) ) is wrong because they substituted incorrectly.So, in reality, ( g(-x) = 2^{x^2 - 4} - 2|1 - x| ), which is not equal to ( g(x) = 2^{x^2 - 4} - 2|x + 1| ), unless ( |1 - x| = |x + 1| ), which isn't true for all ( x ). Therefore, the function is neither even nor odd.Wait, but let me test with specific values to confirm.Take ( x = 1 ):( g(1) = 2^{1 - 4} - 2|1 + 1| = 2^{-3} - 2*2 = 1/8 - 4 = -31/8 )( g(-1) = 2^{1 - 4} - 2| -1 + 1 | = 1/8 - 2*0 = 1/8 )So, ( g(-1) = 1/8 ), which is not equal to ( g(1) = -31/8 ), nor is it equal to ( -g(1) = 31/8 ). So, it's neither even nor odd.Another test, ( x = 2 ):( g(2) = 2^{4 - 4} - 2|2 + 1| = 1 - 6 = -5 )( g(-2) = 2^{4 - 4} - 2| -2 + 1 | = 1 - 2*1 = -1 )Again, ( g(-2) = -1 ) is not equal to ( g(2) = -5 ) or ( -g(2) = 5 ).So, definitely, the function is neither even nor odd.I think the assistant made a mistake in their substitution, which led them to incorrectly conclude the function was even. So, the correct answer is that the function is neither even nor odd.