What is the extreme f(x)=((1))/((3)x^{(3))-400x} ?

The extreme f(x)=((1))/((3)x^{(3))-400x} is Minimum(-20/3 , 9/16000),Maximum(20/3 ,-9/16000)

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extreme f(x)=((1))/((3)x^{(3))-400x}

extreme points

f (x) = \frac{( 1 )}{( 3 ) x ^{(3)} - 400 x}

Minimum (- \frac{20}{3}, \frac{9}{16000}), Maximum (\frac{20}{3}, - \frac{9}{16000})

Solution steps

Find the critical points:

x = - \frac{20}{3}, x = \frac{20}{3}

Domain of

\frac{1}{3 x ^{3} - 400 x} : x < - \frac{20}{3} or - \frac{20}{3} < x < 0 or 0 < x < \frac{20}{3} or x > \frac{20}{3}

Find intervals:

Decreasing

: - \infty < x < - \frac{20}{3},

Decreasing

: - \frac{20}{3} < x < - \frac{20}{3},

Increasing

: - \frac{20}{3} < x < 0,

Increasing

: 0 < x < \frac{20}{3},

Decreasing

: \frac{20}{3} < x < \frac{20}{3},

Decreasing

: \frac{20}{3} < x < \infty

Plug the extreme point

x = - \frac{20}{3}

into

\frac{1}{3 x ^{3} - 400 x} \Rightarrow y = \frac{9}{16000}

Minimum (- \frac{20}{3}, \frac{9}{16000})

Plug the extreme point

x = \frac{20}{3}

into

\frac{1}{3 x ^{3} - 400 x} \Rightarrow y = - \frac{9}{16000}

Maximum (\frac{20}{3}, - \frac{9}{16000})

Minimum (- \frac{20}{3}, \frac{9}{16000}), Maximum (\frac{20}{3}, - \frac{9}{16000})

f (x, y) = x^{2} + x y + 2 x + 4 y + 1

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

f (x) = 2 x^{3} - 21 x^{2} + 60 x, 1 \leq x \leq 6

y^{3} + 6 x^{2} y - 6 x^{2} - 6 y^{2} + 3

What is the extreme f(x)=((1))/((3)x^{(3))-400x} ?
The extreme f(x)=((1))/((3)x^{(3))-400x} is Minimum(-20/3 , 9/16000),Maximum(20/3 ,-9/16000)