g(x)=3x-5, h(x)=x+2, (h-g)(x)

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

☐²

x^☐

√☐

^□√☐

□□

log_□

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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g(x)=3x−5, h(x)=x+2, (h(x)−g(x))(x)

Solution

−2x+7

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Solution steps

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Find (h−g)(x) given g(x)=3x−5, h(x)=x+2

(h−g)(x)=h(x)−g(x)

h(x)=x+2

g(x)=3x−5

=x+2−(3x−5)

Expand x+2−(3x−5): −2x+7

=−2x+7

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

g(x)=3x-5, h(x)=x+2, (h-g)(x)

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