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f(x)=2x+3, g(x)=x^2-5, f+g

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

☐²

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^□√☐

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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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f(x)=2x+3, g(x)=x²−5, f(x)+g(x)

Solution

x²+2x−2

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Find f+g given f(x)=2x+3, g(x)=x²−5

f+g=f(x)+g(x)

f(x)=2x+3

g(x)=x²−5

=2x+3+x²−5

Simplify 2x+3+x²−5: x²+2x−2

=x²+2x−2

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

f(x)=2x+3, g(x)=x^2-5, f+g

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