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

extrempunkte y= x/(x^2-6x+8)

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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 y=xx²−6x+8

Solution

Minimum(−2√2, −3−2√22 ), Maximum(2√2, −2√2+32 )

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Find using the First Derivative Test

Find using the First Derivative Test

Find using the Second Derivative Test

One step at a time

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)=−x²+8(x²−6x+8)²

Find intervals: Decreasing:−∞ <x<−2√2, Increasing:−2√2<x<2, Increasing:2<x<2√2, Decreasing:2√2<x<4, Decreasing:4<x<∞

Plug x=−2√2 into xx²−6x+8 : −3−2√22

Minimum(−2√2, −3−2√22 )

Plug x=2√2 into xx²−6x+8 : −2√2+32

Maximum(2√2, −2√2+32 )

Minimum(−2√2, −3−2√22 ), Maximum(2√2, −2√2+32 )

Solve instead

domain xx²−6x+8

range xx²−6x+8

inverse xx²−6x+8

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