Question

a function f is a... function (polynomial function g is defined by ( g(x) = x^2(9 - x) ). compare the maximum values and the end behavior of the functions ( f ) and ( g ). compute the maximum value of ( f ) and ( g ) when ( x > 0 ). the maximum value of ( f(x) ) is 100 which is less than the maximum value of ( g(x) ) which is near □. (graph of ( f ) below.)

Explanation:

Step1: Rewrite the function

Given \( g(x) = x^2(9 - x) \), expand it: \( g(x)=9x^2 - x^3 \).

Step2: Find the derivative

To find the maximum, take the derivative \( g'(x) \). Using the power rule, \( g'(x)=18x - 3x^2 \).

Step3: Find critical points

Set \( g'(x) = 0 \): \( 18x - 3x^2 = 0 \). Factor out \( 3x \): \( 3x(6 - x)=0 \). So critical points are \( x = 0 \) or \( x = 6 \). Since \( x>0 \), consider \( x = 6 \).

Step4: Evaluate \( g(6) \)

Substitute \( x = 6 \) into \( g(x) \): \( g(6)=6^2(9 - 6)=36\times3 = 108 \). Wait, but the problem says "the maximum value of \( g(x) \) which is near" – wait, maybe miscalculation? Wait, no, \( g(x)=9x^2 - x^3 \), at \( x = 6 \), \( 9*(36)-216=324 - 216 = 108 \). But the first box had 108, then "less than" – maybe the other function? Wait, the image is unclear, but assuming we need to find the maximum of \( g(x)=x^2(9 - x) \) for \( x>0 \), the critical point at \( x = 6 \) gives \( g(6)=108 \). Wait, but maybe the question is comparing with another function. But from the given, if we compute \( g(x) \)'s maximum:

Wait, let's recheck. \( g(x)= -x^3 + 9x^2 \). Derivative \( g'(x)= -3x^2 + 18x = -3x(x - 6) \). So critical points at \( x = 0 \) and \( x = 6 \). For \( x>0 \), test \( x = 6 \). Second derivative: \( g''(x)= -6x + 18 \). At \( x = 6 \), \( g''(6)= -36 + 18 = -18 < 0 \), so it's a local maximum. So \( g(6)=6^2(9 - 6)=36*3 = 108 \). So the maximum value is 108? But the first box has 108, then "less than" – maybe the other function's maximum is more? But based on the given, if we need to find the maximum of \( g(x) \) for \( x>0 \), it's 108.

Answer:

108