Question

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

Functions
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asked 2021-05-18
Find the absolute max and min values at the indicated interval
\(f(x)=2x^{3}-x^{2}-4x+10, [-1,0]\)

Answers (1)

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.

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