What is the extreme y=(25/32 x^3+6400)/x ?

The extreme y=(25/32 x^3+6400)/x is Minimum(16,600)

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extreme y=(25/32 x^3+6400)/x

extreme points

y = \frac{\frac{25}{32} x ^{3} + 6400}{x}

Minimum (16, 600)

Solution steps

f^{'} (x) = \frac{25 ( x ^{3} - 4096 )}{16 x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 16,

Increasing

: 16 < x < \infty

Plug

x = 16

into

\frac{\frac{25}{32} x ^{3} + 6400}{x} : 600

Minimum (16, 600)

f (x) = x ∣ x ∣

E (x) = x^{4} - 8 x + 3 x^{2} + x m - 5 - 11

f (x, y) = y^{3} + (6 x^{2}) y - 6 x^{2} - 6 y^{2} + 3

x^{5}

f (x) = x^{2} - 8 x + 15

What is the extreme y=(25/32 x^3+6400)/x ?
The extreme y=(25/32 x^3+6400)/x is Minimum(16,600)