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) )? a 2 b ( \frac{7}{2} ) c ( \frac{1}{2} ) d 42

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $g(x)=\frac{f(x)}{x}$, then $g^{\prime}(x)=\frac{f^{\prime}(x)\cdot x - f(x)\cdot1}{x^{2}}$.

Step2: Substitute $x = 9$

We know that $f(9)=18$ and $f^{\prime}(9)=7$. Substitute these values into the formula for $g^{\prime}(x)$:
$g^{\prime}(9)=\frac{f^{\prime}(9)\cdot9 - f(9)\cdot1}{9^{2}}$.

Step3: Calculate the numerator

$f^{\prime}(9)\cdot9 - f(9)\cdot1=7\times9 - 18=63 - 18 = 45$.

Step4: Calculate the denominator

$9^{2}=81$.

Step5: Find the value of $g^{\prime}(9)$

$g^{\prime}(9)=\frac{45}{81}=\frac{5}{9}$.

Answer:

$\frac{5}{9}$