Find the absolute max and min values at the indicated interval f(x)=2x3−x2−4x+10,[−1,0]

Question

Find the absolute max and min values at the indicated interval  f(x)=2x3−x2−4x+10,[−1,0]

Raheem Donnelly · Accepted Answer

Given, f(x)=2x3−x2−4x+10,[−1,0] Absolute maximum and absolute minimum values of a function exist at a point where its first derivative is zero or at the end points of the given interval. So first we find the derivative of the given function: fprime(x)=2⋅3x2−2x−4+0 ⇒fprime(x)=6x2−2x−4 Now fprime(x)=0 ⇒6x2−2x−4=0 ⇒6x2−6x+4x−4=0 ⇒6x(x−1)+4(x−1)=0 ⇒(x−1)(6x+4)=0 ⇒x−1=0 or 6x+4=0 ⇒x=1 or x=−23 Now computing the value of the function at these points and at the end points of the interval: at x=−1,f(−1)=f(x)=2(−1)3−(−1)2−4(−1)+10=11 at x=0,f(0)=f(x)=2(0)3−(0)2−4(0)+10=10 at x=−23,f(−23)=f(x)=2(−23)3−(−23)2−4(−23)+10=12.8148 We didn&#039;t find the value of f(1), because 1 does not belongs to the interval [-1,0]. Hence absolute maximum value is 12.8148 and absolute minimum value is 10.

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Find the absolute max and min values at the indicated interval f(x)=2x^{3}-x^{2}-4x+10, \left[-1,0\right]

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