Question

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

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

2021-05-19

Given,
$$f(x)=2x^{3}-x^{2}-4x+10, [-1,0]$$
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:
$$f'(x)=2 \cdot3x^{2}-2x-4+0$$
$$\Rightarrow f'(x)=6x^{2}-2x-4$$
Now $$f'(x)=0$$
$$\Rightarrow 6x^{2}-2x-4=0$$
$$\Rightarrow 6x^{2}-6x+4x-4=0$$
$$\Rightarrow 6x(x-1)+4(x-1)=0$$
$$\Rightarrow (x-1)(6x+4)=0$$
$$\Rightarrow x-1=0\ or\ 6x+4 = 0$$
$$\Rightarrow x=1\ or\ x=-\frac{2}{3}$$
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 $$x=-\frac{2}{3}, f(-\frac{2}{3})=f(x)=2(-\frac{2}{3})^{3}-(-\frac{2}{3})^{2}-4(-\frac{2}{3})+10=12.8148$$
We didn'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.