inverse of sqrt(x+3)

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

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ddx

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

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normal of √x+1, at x=1

Solution

y=−2√2x+3√2

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מצא את נקודת ההשקה: (1, √2)

y=√x+1 חשב את השיפוע של: 12√x+1

Compute the slope at x=1: m=12√2

חשב את השיפוע של הניצב: m_p=−2√2

(1, √2) שעובר דרך m=−2√2 מצא את הישר עם שיפוע: y=−2√2x+3√2

y=−2√2x+3√2

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היי, לשאול שאלה לגבי normal of √x+1,at x=1

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

inverse of sqrt(x+3)

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