Consider the function f(x)=4-6x^{2}, \ \ \ -4\le x\le 2. The

Mary Keefe

Mary Keefe

Answered question

2021-12-10

Consider the function f(x)=46x2,   4x2.
The absolute maximum value is B
and this occurs at x= B
The absolute minimum value is B
and this occurs at x= B

Answer & Explanation

ramirezhereva

ramirezhereva

Beginner2021-12-11Added 28 answers

Step 1
Given:
The function
f(x)=46x2,4x2
Step 2
First calculate the critical point by differentiate f(x)=46x2, and equate it to zero.
f(x)=ddx(46x2)
=-12x
Set f'(x)=0
-12x=0
x=0
Thus the critical point is x=0
At x=0 the value of f(x)=46x2 is
f(0)=46(0)2
f(0)=4...(1)
Step 3
Now find the value of f(x)=46x2 at the endpoints 4x2
At x=-4
f(4)=46(4)2
=4-96
f(-4)=-92...(2)
At x=2
f(2)=46(2)2
=4-24
f(-2)=-20...(3)
By comparing (1),(2), and (3)
The absolute maximum value is 4 and this occur at x=0
The absolute minimum value is -92 and this occur at x=-4

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