What is the extreme f(x)= 4/3 x^3+4x^2-252x+7 ?

The extreme f(x)= 4/3 x^3+4x^2-252x+7 is Maximum(-9,1627),Minimum(7,-3311/3)

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extreme f(x)= 4/3 x^3+4x^2-252x+7

extreme points

f (x) = \frac{4}{3} x^{3} + 4 x^{2} - 252 x + 7

Maximum (- 9, 1627), Minimum (7, - \frac{3311}{3})

Solution steps

f^{'} (x) = 4 x^{2} + 8 x - 252

Find intervals:

Increasing

: - \infty < x < - 9,

Decreasing

: - 9 < x < 7,

Increasing

: 7 < x < \infty

Plug

x = - 9

into

\frac{4}{3} x^{3} + 4 x^{2} - 252 x + 7 : 1627

Maximum (- 9, 1627)

Plug

x = 7

into

\frac{4}{3} x^{3} + 4 x^{2} - 252 x + 7 : - \frac{3311}{3}

Minimum (7, - \frac{3311}{3})

Maximum (- 9, 1627), Minimum (7, - \frac{3311}{3})

f (x) = 4 x - 9 x^{\frac{4}{9}}

f (x, y) = 3 x^{3} - 9 x + 9 x y^{2}

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

f (x, y) = 6 x + 3 y - 7

- 2 x^{2} + 3 x - 1

What is the extreme f(x)= 4/3 x^3+4x^2-252x+7 ?
The extreme f(x)= 4/3 x^3+4x^2-252x+7 is Maximum(-9,1627),Minimum(7,-3311/3)