What is the extreme f(x)=8x^2-6x^4 ?

The extreme f(x)=8x^2-6x^4 is Maximum(-sqrt(2/3), 8/3),Minimum(0,0),Maximum(sqrt(2/3), 8/3)

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extreme f(x)=8x^2-6x^4

extreme points

f (x) = 8 x^{2} - 6 x^{4}

Maximum (- \frac{2}{3}, \frac{8}{3}), Minimum (0, 0), Maximum (\frac{2}{3}, \frac{8}{3})

Solution steps

f^{'} (x) = 16 x - 24 x^{3}

Find intervals:

Increasing

: - \infty < x < - \frac{2}{3},

Decreasing

: - \frac{2}{3} < x < 0,

Increasing

: 0 < x < \frac{2}{3},

Decreasing

: \frac{2}{3} < x < \infty

Plug

x = - \frac{2}{3}

into

8 x^{2} - 6 x^{4} : \frac{8}{3}

Maximum (- \frac{2}{3}, \frac{8}{3})

Plug

x = 0

into

8 x^{2} - 6 x^{4} : 0

Minimum (0, 0)

Plug

x = \frac{2}{3}

into

8 x^{2} - 6 x^{4} : \frac{8}{3}

Maximum (\frac{2}{3}, \frac{8}{3})

Maximum (- \frac{2}{3}, \frac{8}{3}), Minimum (0, 0), Maximum (\frac{2}{3}, \frac{8}{3})

T (x, y) = 27 - 8 x^{2} - 9 y^{2}

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

f (x) = x^{5} - 5 x + 6

f (x) = 2 x^{2} + 3 x y + 4 y^{2} + 2 x - 10 y

f (x) = x^{2} + 4 x + 9

What is the extreme f(x)=8x^2-6x^4 ?
The extreme f(x)=8x^2-6x^4 is Maximum(-sqrt(2/3), 8/3),Minimum(0,0),Maximum(sqrt(2/3), 8/3)