What is the extreme f(x)=7-(7+5x)^{2/5} ?

The extreme f(x)=7-(7+5x)^{2/5} is Maximum(-7/5 ,7)

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extreme f(x)=7-(7+5x)^{2/5}

extreme points

f (x) = 7 - (7 + 5 x)^{\frac{2}{5}}

Maximum (- \frac{7}{5}, 7)

Solution steps

f^{'} (x) = - \frac{2}{( 5 x + 7 ) ^{\frac{3}{5}}}

Find intervals:

Increasing

: - \infty < x < - \frac{7}{5},

Decreasing

: - \frac{7}{5} < x < \infty

Plug

x = - \frac{7}{5}

into

7 - (7 + 5 x)^{\frac{2}{5}} : 7

Maximum (- \frac{7}{5}, 7)

f (x) = \frac{x ^{5}}{5} - \frac{4 x ^{4}}{4} + 2

f (x) = \frac{x ^{5}}{5} - \frac{4 x ^{4}}{4} + 6

f (x, y) = 3 x^{2} - 4 x y + 4 y^{2} - 4 x + 8 y + 4

3 - x

f (x) = 2 x^{3} - 18 x^{2} + 48 x - 7

What is the extreme f(x)=7-(7+5x)^{2/5} ?
The extreme f(x)=7-(7+5x)^{2/5} is Maximum(-7/5 ,7)