What is the critical (7x+6)/(5x^{3/5)} ?

The critical (7x+6)/(5x^{3/5)} is x= 9/7

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critical (7x+6)/(5x^{3/5)}

critical points

\frac{7 x + 6}{5 x ^{\frac{3}{5}}}

x = \frac{9}{7}

Solution steps

\frac{7 x + 6}{5 x ^{\frac{3}{5}}}

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{9}{7}, x = 0

Identify critical points not in

f (x)

domain

Domain of

\frac{7 x + 6}{5 x ^{\frac{3}{5}}} : x < 0 or x > 0

\frac{7 x + 6}{5 x ^{\frac{3}{5}}}

is not defined at

x = 0

, therefore:

x = \frac{9}{7}

2 x^{2}

(- 4, 5), (- 8, 5)

2 x + y = 5

\frac{( e ^{x} )}{1 + 5 e ^{x}}

y = x + 2

What is the critical (7x+6)/(5x^{3/5)} ?
The critical (7x+6)/(5x^{3/5)} is x= 9/7