absolute puntos extremos f(x)=x^3

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

☐²

x^☐

√☐

^□√☐

□□

log_□

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∫

ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

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perpendicular y=−5x+2, x=1

Solution

y=15 x−165

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Solution steps

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Find the line y=mx+b perpendicular to y=−5x+2 that passes through x=1

Find the perpendicular point: (1, −3)

Find the slope of y=−5x+2: m=−5

Compute the slope of the perpendicular line: m_p=15

Find the line with slope m=15 and passing through (1, −3): y=15 x−165

y=15 x−165

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

absolute puntos extremos f(x)=x^3

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