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

The extreme f(x)=-x^3+x^2+x+6 is Minimum(-1/3 , 157/27),Maximum(1,7)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

extreme

Minimum (- \frac{1}{3}, \frac{157}{27}), Maximum (1, 7)

Solution steps

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

Find intervals:

Decreasing

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

Increasing

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

Decreasing

: 1 < x < \infty

Plug

x = - \frac{1}{3}

into

- x^{3} + x^{2} + x + 6 : \frac{157}{27}

Minimum (- \frac{1}{3}, \frac{157}{27})

Plug

x = 1

into

- x^{3} + x^{2} + x + 6 : 7

Maximum (1, 7)

Minimum (- \frac{1}{3}, \frac{157}{27}), Maximum (1, 7)

What is the extreme f(x)=-x^3+x^2+x+6 ?
The extreme f(x)=-x^3+x^2+x+6 is Minimum(-1/3 , 157/27),Maximum(1,7)