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

The absolute minimum value of the function $f\left(x\right)=9{x}^{2}-18x+3$ over the interval [0,10]
(1)6
(2)-6
(3)3
(4)4
(5)1
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Step 1
The absolute minimum value of the function
$f\left(x\right)=9{x}^{2}-18x+3$ over the interval [0,10]
Step 2
The absolute minimum value of the function
$f\left(x\right)=9{x}^{2}-18x+3$ 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)
$fx\right)=\frac{d}{dx}\left(18x-18\right)=18$ which is positive, so function has minimum value.
Step 3
Put x=1 in f(x)
$f\left(1\right)=9{\left(1\right)}^{2}-18\left(1\right)+3$
f(1)=9-18+3
f(1)=-9+3
f(1)=-6
hence minimum absolute value of function is −6