Question

find the absolute maximum and absolute minimum values of f on the given interval.
$f(x)=2x^{3}-3x^{2}-72x + 8$, $-4,5$

Explanation:

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

Using the power rule \((x^n)^\prime=nx^{n - 1}\), for \(f(x)=2x^{3}-3x^{2}-72x + 8\), we have \(f^\prime(x)=6x^{2}-6x - 72\).

Step2: Find the critical points

Set \(f^\prime(x)=0\), so \(6x^{2}-6x - 72 = 0\). Divide through by \(6\) to get \(x^{2}-x - 12=0\).
Factor the quadratic: \((x - 4)(x+ 3)=0\).
Solving \(x - 4=0\) gives \(x = 4\), and solving \(x + 3=0\) gives \(x=-3\). Both \(x=-3\) and \(x = 4\) are in the interval \([-4,5]\).

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

• For \(x=-4\): \(f(-4)=2(-4)^{3}-3(-4)^{2}-72(-4)+8=2(-64)-3(16)+288 + 8=-128-48 + 288+8=128\).
• For \(x=-3\): \(f(-3)=2(-3)^{3}-3(-3)^{2}-72(-3)+8=2(-27)-3(9)+216 + 8=-54-27+216 + 8=143\).
• For \(x = 4\): \(f(4)=2(4)^{3}-3(4)^{2}-72(4)+8=2(64)-3(16)-288 + 8=128-48-288 + 8=-200\).
• For \(x = 5\): \(f(5)=2(5)^{3}-3(5)^{2}-72(5)+8=2(125)-3(25)-360 + 8=250-75-360 + 8=-177\).

Answer:

The absolute maximum value of \(f(x)\) on the interval \([-4,5]\) is \(143\) (at \(x=-3\)), and the absolute minimum value is \(-200\) (at \(x = 4\)).