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

extreme points of f(x)=sin(3x)

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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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extreme points f(x)=sin(3x)

Solution

Maximum(π6 +2π3 n, 1), Minimum(π2 +2π3 n, −1)

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Solve by:

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.

Find the critical points: x=π6 +2π3 n, x=π2 +2π3 n

Domain of sin(3x) : −∞ <x<∞

Find intervals: Increasing:2π3 n≤ x<π6 +2π3 n, Decreasing:π6 +2π3 n<x<π2 +2π3 n, Increasing:π2 +2π3 n<x<2π3 +2π3 n

Plug the extreme point x=π6 +2π3 n into sin(3x) ⇒ y=1

Maximum(π6 +2π3 n, 1)

Plug the extreme point x=π2 +2π3 n into sin(3x) ⇒ y=−1

Minimum(π2 +2π3 n, −1)

Maximum(π6 +2π3 n, 1), Minimum(π2 +2π3 n, −1)

Solve instead

domain sin(3x)

range sin(3x)

periodicity sin(3x)

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