What is the extreme f(x)=x^3+2x^2+x-6 ?

The extreme f(x)=x^3+2x^2+x-6 is Maximum(-1,-6),Minimum(-1/3 ,-166/27)

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extreme f(x)=x^3+2x^2+x-6

extreme

Maximum (- 1, - 6), Minimum (- \frac{1}{3}, - \frac{166}{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 - 6 : - 6

Maximum (- 1, - 6)

Plug

x = - \frac{1}{3}

into

x^{3} + 2 x^{2} + x - 6 : - \frac{166}{27}

Minimum (- \frac{1}{3}, - \frac{166}{27})

Maximum (- 1, - 6), Minimum (- \frac{1}{3}, - \frac{166}{27})

What is the extreme f(x)=x^3+2x^2+x-6 ?
The extreme f(x)=x^3+2x^2+x-6 is Maximum(-1,-6),Minimum(-1/3 ,-166/27)