What is the extreme y=\sqrt[3]{5x^3+5} ?

The extreme y=\sqrt[3]{5x^3+5} is Saddle(0,\sqrt[3]{5})

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extreme y=\sqrt[3]{5x^3+5}

extreme points

y = 3 5 x^{3} + 5

Saddle (0, 35)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

3 5 x^{3} + 5 : 35

Saddle (0, 35)

f (x) = 6 x e^{- 0.5 x}

f (x) = 4 (x - 3)^{2} (x + 1)^{5} (x - 9)

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

f (x) = x^{3} - 12 x^{2} - 27 x + 5, - 2 \leq x \leq 0

f (x) = 4 - 3 x, - 1 \leq x \leq 2

What is the extreme y=\sqrt[3]{5x^3+5} ?
The extreme y=\sqrt[3]{5x^3+5} is Saddle(0,\sqrt[3]{5})