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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

second derivative of test (x+5)^2

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

x^☐

√☐

^□√☐

□□

log_□

π

θ

∞

∫

ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

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second derivative test (x+5)²

Solution

Minimum(−5, 0)

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Solve by:

Find using the Second Derivative Test

Find using the Second Derivative Test

Find using the First Derivative Test

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Second Derivative Test definition

Suppose that x=c is a critical point of f ′(c) such that f ′(c)=0
and that f ′′(x) is continuous in a region around x=c. Then,
If f ′′(c)<0 then x=c is a local maximum.
If f ′′(c)>0 then x=c is a local minimum.
If f ′′(c)=0 then test failed and x=c can be a local maximum, local minimum or neither.

f ′(x)=2(x+5)

Find intervals: Decreasing:−∞ <x<−5, Increasing:−5<x<∞

f ′′(x)=2

Check the sign of f ′′(x)=2 at each critical point

Check critical point x=−5: Minimum(−5, 0)

Minimum(−5, 0)

Solve instead

domain (x+5)²

range (x+5)²

vertices (x+5)²

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