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

Danelle Albright

Danelle Albright

Answered question

2021-12-10

Consider function f(x)=12x2,5x1.
The absolute maximum value is
and this occurs at x=
The absolute minimum value is
and this occurs at x=

Answer & Explanation

Papilys3q

Papilys3q

Beginner2021-12-11Added 34 answers

Step 1
Consider the function
f(x)=12x2,5x1.
Step 2
Since the given function is a polynomial and so it is continuous everywhere and therefore is continuous on the given interval.
Now compute its first derivative and put it equal to zero to find its critical points
f(x)=12x2
f'(x)=0-4x
f'(x)=-4x
f'(x)=0
4x=0
x=0
The critical point x=0 belong to the given interval.
Step 3
Now we find the function value at the critical points and the end points of the interval.
f(x)=12x2
f(5)=12(5)2
=1-50
f(-5)=-49
f(0)=12(0)2
f(0)=1
f(1)=12(1)2
f(1)=-1
Absolute Maximum value is 1
and this occurs at x=0.
Absolute Minimum value is -49
and this occurs at x=5.

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