What is the extreme y=(x^3)/3+4x^2+15x+7 ?

The extreme y=(x^3)/3+4x^2+15x+7 is Maximum(-5,-29/3),Minimum(-3,-11)

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extreme y=(x^3)/3+4x^2+15x+7

extreme points

y = \frac{x ^{3}}{3} + 4 x^{2} + 15 x + 7

Maximum (- 5, - \frac{29}{3}), Minimum (- 3, - 11)

Solution steps

f^{'} (x) = x^{2} + 8 x + 15

Find intervals:

Increasing

: - \infty < x < - 5,

Decreasing

: - 5 < x < - 3,

Increasing

: - 3 < x < \infty

Plug

x = - 5

into

\frac{x ^{3}}{3} + 4 x^{2} + 15 x + 7 : - \frac{29}{3}

Maximum (- 5, - \frac{29}{3})

Plug

x = - 3

into

\frac{x ^{3}}{3} + 4 x^{2} + 15 x + 7 : - 11

Minimum (- 3, - 11)

Maximum (- 5, - \frac{29}{3}), Minimum (- 3, - 11)

x^{2} - 3 x - 4

f (x) = e^{x} ln (x) y

f (x) = 2 x - 4 x - 7

f (x) = \frac{2 x + 1}{2 x}

f (x) = x^{2} lo g_{10} (4) (x)

What is the extreme y=(x^3)/3+4x^2+15x+7 ?
The extreme y=(x^3)/3+4x^2+15x+7 is Maximum(-5,-29/3),Minimum(-3,-11)