What is the extreme f(x)=4-(8+7x)^{2/7} ?

The extreme f(x)=4-(8+7x)^{2/7} is Maximum(-8/7 ,4)

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extreme f(x)=4-(8+7x)^{2/7}

extreme points

f (x) = 4 - (8 + 7 x)^{\frac{2}{7}}

Maximum (- \frac{8}{7}, 4)

Solution steps

f^{'} (x) = - \frac{2}{( 7 x + 8 ) ^{\frac{5}{7}}}

Find intervals:

Increasing

: - \infty < x < - \frac{8}{7},

Decreasing

: - \frac{8}{7} < x < \infty

Plug

x = - \frac{8}{7}

into

4 - (8 + 7 x)^{\frac{2}{7}} : 4

Maximum (- \frac{8}{7}, 4)

\frac{3}{( x - 1 ) ( x ^{2} - 4 )}

f (x, y) = 2 x^{2} - 8 y^{2} - 4 x + 9 y + 2

f (x) = - 6 x^{2} - 12 x + 18

\frac{2 x + 2}{x - 2}

- x^{2} + 2 x

What is the extreme f(x)=4-(8+7x)^{2/7} ?
The extreme f(x)=4-(8+7x)^{2/7} is Maximum(-8/7 ,4)