What is the extreme f(x)=| 1/16 x^3-4| ?

The extreme f(x)=| 1/16 x^3-4| is Saddle(0,4),Minimum(4,0)

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extreme f(x)=| 1/16 x^3-4|

extreme points

f (x) = \frac{1}{16} x^{3} - 4

Saddle (0, 4), Minimum (4, 0)

Solution steps

Find the critical points:

x = 0, x = 4

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug the extreme point

x = 0

into

\frac{1}{16} x^{3} - 4 \Rightarrow y = 4

Saddle (0, 4)

Plug the extreme point

x = 4

into

\frac{1}{16} x^{3} - 4 \Rightarrow y = 0

Minimum (4, 0)

Saddle (0, 4), Minimum (4, 0)

\frac{x - 5}{4}

f (x, y) = x^{2} + y^{2} - 18 x + 16 y - 7

f (x) = \frac{x - 4}{x ^{2} - 7}

f (x, y) = x^{3} - 48 x y + 64 y^{3}

f (x) = 7 x - 14 cos (x), - 2 \leq x \leq 0

What is the extreme f(x)=| 1/16 x^3-4| ?
The extreme f(x)=| 1/16 x^3-4| is Saddle(0,4),Minimum(4,0)