What is the critical f(x)= 1/2 x^2+8x+7 ?

The critical f(x)= 1/2 x^2+8x+7 is x=-8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

critical f(x)= 1/2 x^2+8x+7

critical points

f (x) = \frac{1}{2} x^{2} + 8 x + 7

x = - 8

Solution steps

\frac{1}{2} x^{2} + 8 x + 7

Find where

f^{'} (x)

is equal to zero or undefined

x = - 8

Identify critical points not in

f (x)

domain

Domain of

\frac{1}{2} x^{2} + 8 x + 7 : - \infty < x < \infty

All critical points are in domain

x = - 8

- 5 cos (6 x)

f (x) = y - 6 x = 42

f (x) = - 3 x^{4} + 24 x^{3} - 48 x^{2}

f (x) = 3 x + 5 - 4

ln (tan (x)) - x^{2}

What is the critical f(x)= 1/2 x^2+8x+7 ?
The critical f(x)= 1/2 x^2+8x+7 is x=-8