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

The extreme f(x)=x^3+2x^2+x-7 is Maximum(-1,-7),Minimum(-1/3 ,-193/27)

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

extreme points

f (x) = x^{3} + 2 x^{2} + x - 7

Maximum (- 1, - 7), Minimum (- \frac{1}{3}, - \frac{193}{27})

Solution steps

f^{'} (x) = 3 x^{2} + 4 x + 1

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < - \frac{1}{3},

Increasing

: - \frac{1}{3} < x < \infty

Plug

x = - 1

into

x^{3} + 2 x^{2} + x - 7 : - 7

Maximum (- 1, - 7)

Plug

x = - \frac{1}{3}

into

x^{3} + 2 x^{2} + x - 7 : - \frac{193}{27}

Minimum (- \frac{1}{3}, - \frac{193}{27})

Maximum (- 1, - 7), Minimum (- \frac{1}{3}, - \frac{193}{27})

f (x) = x^{2} - 4

\frac{x ^{2} + 5 x + 6}{x + 15}

f (x) = x^{2} - 6 x + 13

f (x) = \frac{1}{∣ x ^{2} - 5 x + 6 ∣}

f (x) = \frac{1}{2 x + 4}

What is the extreme f(x)=x^3+2x^2+x-7 ?
The extreme f(x)=x^3+2x^2+x-7 is Maximum(-1,-7),Minimum(-1/3 ,-193/27)