bantu bang cepat ha8udhc.... Question from @omisaputra19

omisaputra19 @omisaputra19

July 2022 1 4 Report

bantu bang cepat ha8udhc

JΟY

${\sf{1.~(f+g)(x)=2x^2-x-1}}$
${\sf{2.~(f-g)(x)=2x^2-3x-1}}$
${\sf{3.~({f}\times{g})(x)=2x^3-4x^2+2x}}$
${\sf{4.~\left(\dfrac{f}{g}\right)(x)=2x}}$
${\sf{5.~({f}\circ{g})(2)=-19}}$

Penjelasan dengan langkah-langkah :

» No. 1 «

Diketahui

${\sf{f(x)=2x^2-2x}}$
${\sf{g(x)=x-1}}$

Ditanya

${\sf{(f+g)(x)\dots?}}$

Jawaban

$\begin{aligned}{\sf{(f+g)(x)}}&={\sf{f(x)+g(x)}}\\&={\sf{(2x^2-2x)+(x-1)}}\\&={\sf{2x^2-2x+x-1}}\\&={\sf{2x^2-x-1}}\end{aligned}$

» No. 2 «

Diketahui

${\sf{f(x)=2x^2-2x}}$
${\sf{g(x)=x-1}}$

Ditanya

${\sf{(f-g)(x)\dots?}}$

Jawaban

$\begin{aligned}{\sf{(f-g)(x)}}&={\sf{f(x)-g(x)}}\\&={\sf{(2x^2-2x)-(x-1)}}\\&={\sf{2x^2-2x-x-1}}\\&={\sf{2x^2-3x-1}}\end{aligned}$

» No. 3 «

Diketahui

${\sf{f(x)=2x^2-2x}}$
${\sf{g(x)=x-1}}$

Ditanya

${\sf{({f}\times{g})(x)\dots?}}$

Jawaban

$\begin{aligned}{\sf{({f}\times{g})(x)}}&={\sf{{f(x)}\times{g(x)}}}\\&={\sf{(2x^2-2x)(x-1)}}\\&={\sf{2x^2(x-1)-2x(x-1)}}\\&={\sf{2x^3-2x^2-2x^2+2x}}\\&={\sf{2x^3-4x^2+2x}}\end{aligned}$

» No. 4 «

Diketahui

${\sf{f(x)=2x^2-2x}}$
${\sf{g(x)=x-1}}$

Ditanya

${\sf{\left(\dfrac{f}{g}\right)(x)\dots?}}$

Jawaban

$\begin{aligned}{\sf{\left(\dfrac{f}{g}\right)(x)}}&={\sf{\frac{f(x)}{g(x)}}}\\&={\sf{\frac{2x^2-2x}{x-1}}}\\&={\sf{\frac{2x\cancel{(x-1)}}{\cancel{x-1}}}}\\&={\sf{2x}}\end{aligned}$

» No. 5 «

Diketahui

${\sf{f(x)=\dfrac{2x+7}{x-7}}}$
${\sf{g(x)=2x+2}}$

Ditanya

${\sf{({f}\circ{g})(x)\dots?}}$

Jawaban

$\begin{aligned}{\sf{({f}\circ{g})(x)}}&={\sf{f(g(x))}}\\&={\sf{\frac{2(2x+2)+7}{(2x+2)-7}}}\\&={\sf{\frac{4x+4+7}{2x+2-7}}}\\&={\sf{\frac{4x+11}{2x-5}}}\\{\sf{({f}\circ{g})(2)}}&={\sf{\frac{4(2)+11}{2(2)-5}}}\\&={\sf{\frac{8+11}{4-5}}}\\&={\sf{\frac{19}{-1}}}\\&={\sf{-19}}\\\end{aligned}$