Question

let ( f ) be the function given by ( f(x) = -x^3 + 3x^2 + 24x ). what is the absolute maximum value of ( f ) on the closed interval ( -6, 6 )?

a ( -6 )

b ( 36 )

c ( 80 )

d ( 180 )

Explanation:

Step1: Find the derivative of \( f(x) \)

To find critical points, we first find the derivative of \( f(x)=-x^{3}+3x^{2}+24x \). Using the power rule, \( f^\prime(x)=-3x^{2}+6x + 24 \).

Step2: Solve \( f^\prime(x) = 0 \) for critical points

Set \( -3x^{2}+6x + 24 = 0 \). Divide both sides by -3: \( x^{2}-2x - 8=0 \). Factor the quadratic: \( (x - 4)(x + 2)=0 \). So critical points are \( x = 4 \) and \( x=-2 \).

Step3: Evaluate \( f(x) \) at critical points and endpoints

• Evaluate at \( x=-6 \): \( f(-6)=-(-6)^{3}+3(-6)^{2}+24(-6)=216 + 108-144 = 180 \)
• Evaluate at \( x=-2 \): \( f(-2)=-(-2)^{3}+3(-2)^{2}+24(-2)=8 + 12-48=-28 \)
• Evaluate at \( x = 4 \): \( f(4)=-(4)^{3}+3(4)^{2}+24(4)=-64 + 48 + 96 = 80 \)
• Evaluate at \( x = 6 \): \( f(6)=-(6)^{3}+3(6)^{2}+24(6)=-216+108 + 144 = 36 \)

Step4: Compare the values

We have \( f(-6)=180 \), \( f(-2)=-28 \), \( f(4)=80 \), \( f(6)=36 \). The largest value is 180.

Answer:

D. 180