Сonsider the function f(x)=2x^{2}-2x+8, 0\le x\le 10. The absolute maximum of

korporasidn

korporasidn

Answered question

2021-12-03

Сonsider the function f(x)=2x22x+8,0x10.
The absolute maximum of f(x) (on the given interval) is at x=
and the absolute minimum of f(x) (on the given interval) is at x=

Answer & Explanation

mylouscrapza

mylouscrapza

Beginner2021-12-04Added 22 answers

Step 1
Given function is:
f(x)=2x22x+8
Now,
First, find the critical points of the given function.
Therefore,
f(x)=df(x)dx
=d(2x22x+8)dx
=4x-2
Then,
Put f'(x)=0
∴⇒4x2=0
4x=2
x=12
Step 2
For absolute values, check the value of the function at critical point and endpoints of the given interval.
Therefore,
f(0)=2×022×0+8=8
f(10)=2×1022×10+8=188
f(12)=2×(12)22×12+8=7.5
Thus,
The absolute maximum of f(x) (on the given interval) is at x=10
The absolute minimum of f(x) (on the given interval) is at x=12

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