f(x)=2x^2+x-3, g(x)=x-1, f*g

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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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f(x)=2x²+x−3, g(x)=x−1, f(x)· g(x)

Solution

2x³−x²−4x+3

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Solution steps

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Find f· g given f(x)=2x²+x−3, g(x)=x−1

f· g=f(x)· g(x)

f(x)=2x²+x−3

g(x)=x−1

=(2x²+x−3)(x−1)

Expand (2x²+x−3)(x−1): 2x³−x²−4x+3

=2x³−x²−4x+3

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

f(x)=2x^2+x-3, g(x)=x-1, f*g

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