What is the extreme f(x)= 1/(x^3-6x^2+1) ?

The extreme f(x)= 1/(x^3-6x^2+1) is Minimum(0,1),Maximum(4,-1/31)

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extreme f(x)= 1/(x^3-6x^2+1)

extreme points

f (x) = \frac{1}{x ^{3} - 6 x ^{2} + 1}

Minimum (0, 1), Maximum (4, - \frac{1}{31})

Solution steps

Find the critical points:

x = 0, x = 4

Domain of

\frac{1}{x ^{3} - 6 x ^{2} + 1} : x < - 0.39542 \dots or - 0.39542 \dots < x < 0.42346 \dots or 0.42346 \dots < x < 5.97196 \dots or x > 5.97196 \dots

Find intervals:

Decreasing

: - \infty < x < - 0.39542 \dots,

Decreasing

: - 0.39542 \dots < x < 0,

Increasing

: 0 < x < 0.42346 \dots,

Increasing

: 0.42346 \dots < x < 4,

Decreasing

: 4 < x < 5.97196 \dots,

Decreasing

: 5.97196 \dots < x < \infty

Plug the extreme point

x = 0

into

\frac{1}{x ^{3} - 6 x ^{2} + 1} \Rightarrow y = 1

Minimum (0, 1)

Plug the extreme point

x = 4

into

\frac{1}{x ^{3} - 6 x ^{2} + 1} \Rightarrow y = - \frac{1}{31}

Maximum (4, - \frac{1}{31})

Minimum (0, 1), Maximum (4, - \frac{1}{31})

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

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

f (x) = \frac{x}{x ^{2} - 9}

f (x) = x ln (x)

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

What is the extreme f(x)= 1/(x^3-6x^2+1) ?
The extreme f(x)= 1/(x^3-6x^2+1) is Minimum(0,1),Maximum(4,-1/31)