Consider the function f(x)=5-3x^{2}, -5\le x\le 1. The absolute maximum value

ZIHLOLEp3

ZIHLOLEp3

Answered question

2021-12-04

Consider the function f(x)=53x2,5x1.
The absolute maximum value is
and this occurs at x=
The absolute minimum value is
and this occurs at x=

Answer & Explanation

Jeffrey Parrish

Jeffrey Parrish

Beginner2021-12-05Added 15 answers

Step 1
The given function is:
f(x)=53x2,5x1
Step 2
To find the critical point of the given function:
f'(x)=0
-6x=0
x=0
Therefore:
f(0)=5
f(-5)=-70 and f(1)=2
Now:
f(-5)f(1)
Hence the given function is increasing function on the interval [-5, 0] and decreasing on the interval [0, 1].
Hence:
The absolute maximum value is 5 and this occurs at x = 0.
The absolue minimum value is -70 and this occurs at x = -5.

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