Question

let f be a differentiable function such that f(9)=18 and f(9)=7. if g is the function defined by g(x)=\frac{f(x)}{sqrt{x}}, what is the value of g(9)?

Explanation:

Step1: Apply quotient - rule for differentiation

The quotient - rule states that if $g(x)=\frac{u(x)}{v(x)}$, then $g^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v^{2}(x)}$. Here, $u = f(x)$ and $v=\sqrt{x}=x^{\frac{1}{2}}$. So $u^{\prime}=f^{\prime}(x)$ and $v^{\prime}=\frac{1}{2}x^{-\frac{1}{2}}$. Then $g^{\prime}(x)=\frac{f^{\prime}(x)\sqrt{x}-f(x)\frac{1}{2\sqrt{x}}}{x}$.

Step2: Substitute $x = 9$

We know that $f(9)=18$ and $f^{\prime}(9)=7$. Substitute $x = 9$ into the formula for $g^{\prime}(x)$:
\[

$$\begin{align*} g^{\prime}(9)&=\frac{f^{\prime}(9)\sqrt{9}-f(9)\frac{1}{2\sqrt{9}}}{9}\\ &=\frac{7\times3 - 18\times\frac{1}{2\times3}}{9}\\ &=\frac{21 - 3}{9}\\ &=\frac{18}{9}\\ & = 2 \end{align*}$$

\]

Answer:

A. 2