what is the inverse of $f(x)=\frac{1}{3}x + 2$?\n$h(x)=\frac{1}{3}x + 2$\n$h(x)=\frac{1}{3}x - 2$\n$h(x)=3x…
what is the inverse of $f(x)=\frac{1}{3}x + 2$?\n$h(x)=\frac{1}{3}x + 2$\n$h(x)=\frac{1}{3}x - 2$\n$h(x)=3x - 2$\n$h(x)=3x - 6$
Answer
Answer:
D. $h(x)=3x - 6$
Explanation:
Step1: Let $y = f(x)$
$y=\frac{1}{3}x + 2$
Step2: Swap $x$ and $y$
$x=\frac{1}{3}y + 2$
Step3: Solve for $y$
Subtract 2 from both sides: $x - 2=\frac{1}{3}y$. Multiply both sides by 3: $y = 3(x - 2)=3x-6$. So the inverse function $h(x)=3x - 6$.