inverse ofr ((x+3)^2}{25}-\frac{(y-4)^2)/9 =1

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

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√☐

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≥	≤	·	÷	`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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critical points f(x)=√x+3

Solution

x=−3

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Solution steps

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√x+3

Critical point definition

Critical points are points where the function is defined and its derivative is zero or undefined

Find where f ′(x) is equal to zero or undefined

x=−3

Identify critical points not in f(x) domain

Domain of √x+3 : x≥ −3

All critical points are in domain

x=−3

Solve instead

domain √x+3

range √x+3

inverse √x+3

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

inverse ofr ((x+3)^2}{25}-\frac{(y-4)^2)/9 =1

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