What is the critical f(x)= 1/2 x^2+8x+3 ?

The critical f(x)= 1/2 x^2+8x+3 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+3

critical points

f (x) = \frac{1}{2} x^{2} + 8 x + 3

x = - 8

Solution steps

\frac{1}{2} x^{2} + 8 x + 3

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 + 3 : - \infty < x < \infty

All critical points are in domain

x = - 8

\frac{1}{3}

f (x) = - 5 x^{2} + 8 x - 4

\frac{x + 1}{x + 7}

x^{3} - x^{4}

y = - 7 x - 51

What is the critical f(x)= 1/2 x^2+8x+3 ?
The critical f(x)= 1/2 x^2+8x+3 is x=-8