What is the extreme f(x)=4x^2-32ln(x) ?

The extreme f(x)=4x^2-32ln(x) is Minimum(2,16-32ln(2))

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extreme f(x)=4x^2-32ln(x)

extreme

Minimum (2, 16 - 32 ln (2))

Solution steps

f^{'} (x) = 8 x - \frac{32}{x}

Find intervals:

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 2

into

4 x^{2} - 32 ln (x) : 16 - 32 ln (2)

Minimum (2, 16 - 32 ln (2))

What is the extreme f(x)=4x^2-32ln(x) ?
The extreme f(x)=4x^2-32ln(x) is Minimum(2,16-32ln(2))