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

first derivative of test 1/(x^2)+1/x

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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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extreme points 1x² +1x

Solution

Minimum(−2, −14 )

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Find using the First Derivative Test

Find using the First Derivative Test

Find using the Second Derivative Test

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First Derivative Test definition

Suppose that x=c is a critical point of f(x) then,
If f ′(x)>0 to the left of x=c and f ′(x)<0 to the right of x=c then x=c is a local maximum.
If f ′(x)<0 to the left of x=c and f ′(x)> 0 to the right of x=c then x=c is a local minimum.
If f ′(x) is the same sign on both sides of x=c then x=c is neither a local maximum nor a local minimum.

f ′(x)=−2x³ −1x²

Find intervals: Decreasing:−∞ <x<−2, Increasing:−2<x<0, Decreasing:0<x<∞

Plug x=−2 into 1x² +1x : −14

Minimum(−2, −14 )

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domain 1x² +1x

range 1x² +1x

inverse 1x² +1x

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