The absolute minimum value of the function f(x)=9x^{2}-18x+3 over the

hrostentsp6 2021-12-04 Answered
The absolute minimum value of the function f(x)=9x218x+3 over the interval [0,10]
(1)6
(2)-6
(3)3
(4)4
(5)1
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Expert Answer

Ryan Willis
Answered 2021-12-05 Author has 15 answers
Step 1
The absolute minimum value of the function
f(x)=9x218x+3 over the interval [0,10]
Step 2
The absolute minimum value of the function
f(x)=9x218x+3 over the interval [0,10]
In first step we need to find first derivative of f(x) and put it equal to 0
f'(x)=18x−18
put f'(x)=0
18x-18=0
18x=18
x=1
find f"(x)
fx)=ddx(18x18)=18 which is positive, so function has minimum value.
Step 3
Put x=1 in f(x)
f(1)=9(1)218(1)+3
f(1)=9-18+3
f(1)=-9+3
f(1)=-6
hence minimum absolute value of function is −6
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