What is the extreme f(x)=5+(8+7x)^{2/7} ?

The extreme f(x)=5+(8+7x)^{2/7} is Minimum(-8/7 ,5)

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extreme f(x)=5+(8+7x)^{2/7}

extreme points

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

Minimum (- \frac{8}{7}, 5)

Solution steps

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

Find intervals:

Decreasing

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

Increasing

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

Plug

x = - \frac{8}{7}

into

5 + (8 + 7 x)^{\frac{2}{7}} : 5

Minimum (- \frac{8}{7}, 5)

F

y = \frac{6 x}{x ^{2} + 4}

f (x) = 4 - 3 x - x^{2}

f (x) = f (x) = x^{3} - 9 x^{2} + 15 x + 4

f (x) = \frac{x ^{2} + 81}{4 x}, 1 \leq x \leq 12

What is the extreme f(x)=5+(8+7x)^{2/7} ?
The extreme f(x)=5+(8+7x)^{2/7} is Minimum(-8/7 ,5)