What is the extreme f(x)=2x^2-16ln(x) ?

The extreme f(x)=2x^2-16ln(x) is Minimum(2,8-16ln(2))

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extreme f(x)=2x^2-16ln(x)

extreme points

f (x) = 2 x^{2} - 16 ln (x)

Minimum (2, 8 - 16 ln (2))

Solution steps

f^{'} (x) = 4 x - \frac{16}{x}

Find intervals:

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 2

into

2 x^{2} - 16 ln (x) : 8 - 16 ln (2)

Minimum (2, 8 - 16 ln (2))

f (x, y) = x 1 + y^{3}

f (x) = \frac{1 + x}{1 + x ^{2}}

f (x) = x^{2} y + 2 x y - y^{2} - 3 y

f (x) = x^{3} - 3 x + 2, (- 2, 2)

f (x) = 288 y^{2} + x^{2} - x^{2} y

What is the extreme f(x)=2x^2-16ln(x) ?
The extreme f(x)=2x^2-16ln(x) is Minimum(2,8-16ln(2))