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Popular >

extreme f(x)=x^2+4x+16

Solution

extreme points

f (x) = x^{2} + 4 x + 16

Solution

Minimum (- 2, 12)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < \infty

Plug

x = - 2

into

x^{2} + 4 x + 16 : 12

Minimum (- 2, 12)

Graph

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Frequently Asked Questions (FAQ)

What is the extreme f(x)=x^2+4x+16 ?
The extreme f(x)=x^2+4x+16 is Minimum(-2,12)