What is the extreme y=\sqrt[3]{x^2+3x-10} ?

The extreme y=\sqrt[3]{x^2+3x-10} is Minimum(-3/2 ,-\sqrt[3]{49/4})

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extreme y=\sqrt[3]{x^2+3x-10}

extreme points

Solution steps

f^{'} (x) = \frac{2 x + 3}{3 ( x ^{2} + 3 x - 10 ) ^{\frac{2}{3}}}

Find intervals:

Decreasing

: - \infty < x < - 5,

Decreasing

: - 5 < x < - \frac{3}{2},

Increasing

: - \frac{3}{2} < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - \frac{3}{2}

into

f (x, y) = x y - 2 x - 4 y

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

f (x) = x^{3} - 4 x^{2} - 3 x + 6

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

x^{3} - 6

What is the extreme y=\sqrt[3]{x^2+3x-10} ?
The extreme y=\sqrt[3]{x^2+3x-10} is Minimum(-3/2 ,-\sqrt[3]{49/4})