What is the extreme f(x)=-2x^2ln(x)+17x^2 ?

The extreme f(x)=-2x^2ln(x)+17x^2 is Maximum(e^8,e^{16})

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extreme f(x)=-2x^2ln(x)+17x^2

extreme points

f (x) = - 2 x^{2} ln (x) + 17 x^{2}

Maximum (e^{8}, e^{16})

Solution steps

f^{'} (x) = - 4 x (ln (x) - 8)

Find intervals:

Increasing

: 0 < x < e^{8},

Decreasing

: e^{8} < x < \infty

Plug

x = e^{8}

into

- 2 x^{2} ln (x) + 17 x^{2} : e^{16}

Maximum (e^{8}, e^{16})

∣ x + 4 ∣

f (x) = \frac{2 x ^{2} + 3}{x ^{2} - 1}

f (x) = 4 x + 8 cos (x)

f (y) = x^{2} + y^{2} - 20 x + 16 y - 9

f (x, y) = x^{4} + y^{4} + 4 x y

What is the extreme f(x)=-2x^2ln(x)+17x^2 ?
The extreme f(x)=-2x^2ln(x)+17x^2 is Maximum(e^8,e^{16})