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

inverse of laplace 2/(x^2+8x+17)

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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

simplify solve for inverse tangent line

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second derivative test 2x²+6x

Solution

Maximum(−√3, −4√3), Minimum(√3, 4√3)

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Solve by:

Find using the Second Derivative Test

Find using the Second Derivative Test

Find using the First Derivative Test

One step at a time

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)=2x²−6x²

Find intervals: Increasing:−∞ <x<−√3, Decreasing:−√3<x<0, Decreasing:0<x<√3, Increasing:√3<x<∞

f ′′(x)=12x³

Check the sign of f ′′(x)=12x³ at each critical point

Check critical point x=−√3: Maximum(−√3, −4√3)

Check critical point x=√3: Minimum(√3, 4√3)

Maximum(−√3, −4√3), Minimum(√3, 4√3)

Solve instead

domain 2x²+6x

range 2x²+6x

inverse 2x²+6x

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