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

second derivative of test 4-x^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 4−x²

Solution

Maximum(0, 4)

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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)=−2x

Find intervals: Increasing:−∞ <x<0, Decreasing:0<x<∞

f ′′(x)=−2

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

Check critical point x=0: Maximum(0, 4)

Maximum(0, 4)

Solve instead

domain 4−x²

range 4−x²

vertices 4−x²

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