Question

If \(\mathrm{f}(x)=3 x+10, \mathrm{~g}(x)=x^2-1\), then \((\mathrm{fog})^{-1}(x)=\)

• A. \((\frac{x-7}{3})\)
• B. \((\frac{x-7}{3})^{\frac{1}{2}}\)
• C. \((\frac{x-7}{3})^{\frac{1}{3}}\)
• D. \((\frac{3}{x-7})^{\frac{3}{2}}\)

Hint:

Form fog(x), then solve y = fog(x) for x.

Answer 1

The Correct Option is B

Step 1: Compose
\((f\circ g)(x) = f(g(x)) = 3(x^2 - 1) + 10 = 3x^2 + 7\).

Step 2: Solve for x
Let \(y = 3x^2 + 7\). Then \(x^2 = \dfrac{y - 7}{3}\) and \(x = \left(\dfrac{y - 7}{3}\right)^{1/2}\).

Step 3: Inverse
Interchange the variables: \((f\circ g)^{-1}(x) = \left(\dfrac{x - 7}{3}\right)^{1/2}\).

Final Answer:
The inverse is \(\left(\dfrac{x-7}{3}\right)^{1/2}\). \[ \boxed{\text{(B) }\left(\dfrac{x-7}{3}\right)^{\frac{1}{2}}} \]