Question

consider the function $f(x)=2x^{3}+12x^{2}-30x + 4$, $-5leq xleq2$. this function has an absolute minimum value equal to and an absolute maximum value equal to

Explanation:

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

The derivative \(f^\prime(x)\) of \(f(x)=2x^{3}+12x^{2}-30x + 4\) is \(f^\prime(x)=6x^{2}+24x - 30\). Factor out \(6\): \(f^\prime(x)=6(x^{2}+4x - 5)=6(x + 5)(x - 1)\).

Step2: Find the critical points

Set \(f^\prime(x)=0\). Using the zero - product property \(6(x + 5)(x - 1)=0\), we get \(x=-5\) or \(x = 1\). Both \(x=-5\) and \(x = 1\) are in the interval \([-5,2]\).

Step3: Evaluate the function at the critical points and endpoints

• For \(x=-5\): \(f(-5)=2(-5)^{3}+12(-5)^{2}-30(-5)+4=2(-125)+12(25)+150 + 4=-250+300+150 + 4=204\).
• For \(x = 1\): \(f(1)=2(1)^{3}+12(1)^{2}-30(1)+4=2 + 12-30 + 4=-12\).
• For \(x = 2\): \(f(2)=2(2)^{3}+12(2)^{2}-30(2)+4=2(8)+12(4)-60 + 4=16+48-60 + 4=8\).

Answer:

The absolute minimum value is \(-12\) and the absolute maximum value is \(204\).