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

first derivative of test 1/(9+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

simplify solve for inverse tangent line

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extreme points 19+x²

Solution

Maximum(0, 19 )

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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)=−2x(9+x²)²

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

Plug x=0 into 19+x² : 19

Maximum(0, 19 )

Solve instead

domain 19+x²

inverse 19+x²

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