What is the extreme f(x)= 4/3 x^3+7x^2-8x-42 ?

The extreme f(x)= 4/3 x^3+7x^2-8x-42 is Maximum(-4, 50/3),Minimum(1/2 ,-529/12)

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extreme f(x)= 4/3 x^3+7x^2-8x-42

extreme points

f (x) = \frac{4}{3} x^{3} + 7 x^{2} - 8 x - 42

Maximum (- 4, \frac{50}{3}), Minimum (\frac{1}{2}, - \frac{529}{12})

Solution steps

f^{'} (x) = 4 x^{2} + 14 x - 8

Find intervals:

Increasing

: - \infty < x < - 4,

Decreasing

: - 4 < x < \frac{1}{2},

Increasing

: \frac{1}{2} < x < \infty

Plug

x = - 4

into

\frac{4}{3} x^{3} + 7 x^{2} - 8 x - 42 : \frac{50}{3}

Maximum (- 4, \frac{50}{3})

Plug

x = \frac{1}{2}

into

\frac{4}{3} x^{3} + 7 x^{2} - 8 x - 42 : - \frac{529}{12}

Minimum (\frac{1}{2}, - \frac{529}{12})

Maximum (- 4, \frac{50}{3}), Minimum (\frac{1}{2}, - \frac{529}{12})

f (x) = 2 x 16 - x^{2}

f (x) = 16 - x^{2} - y^{2}

f (x, y) = 25 x^{2} + 4 y^{2} + 4

f (x) = - 4 x^{3} + 12 x^{2} + 4 x - 12

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

What is the extreme f(x)= 4/3 x^3+7x^2-8x-42 ?
The extreme f(x)= 4/3 x^3+7x^2-8x-42 is Maximum(-4, 50/3),Minimum(1/2 ,-529/12)