Q. 46 - Calculus

Question

Use a sign chart for $f\text{'}\text{'}$ to determine the intervals on which each function f in Exercises 41–52 is concave up or concave down, and identify the locations of any inflection points. Then verify your algebraic answers with graphs from a calculator or graphing utility 

$f\left(x\right)=\frac{\sqrt{x}}{x-2}$

Step-by-Step Solution

Answer

Inflection points are $x=\frac{2(2\sqrt{3}-3)}{3}$, concave up on $\left(0,\frac{2(2\sqrt{3}-3)}{3}\right),(2,\infty )\text{ and concave down on }\left(\frac{2(2\sqrt{3}-3)}{3},2\right)\text{. }$

Step 1. Given Information.

The given function is $f\left(x\right)=\frac{\sqrt{x}}{x-2}$.

Step 2. Second derivative.

On differentiating,

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d}{dx}\frac{\sqrt{x}}{x-2} \\ =\frac{\frac{(x-2)}{2\sqrt{x}}-\sqrt{x}}{(x-2{)}^2} \\ =-\frac{x+2}{2(x-2{)}^2\sqrt{x}} \\ {f}^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}-\frac{x+2}{2(x-2{)}^2\sqrt{x}} \\ =\frac{3x^2+12x-4}{4(x-2{)}^3{x}^{\frac{3}{2}}} \end{gathered}$$

Step 3. Sign chart.

Now,

$f^{\text{'}\text{'}}\left(x\right)\text{ is undefined on }x=0,2$

Therefore, the sign chart will be,

Inflection point is $x=\frac{2(2\sqrt{3}-3)}{3}$.

Concave up on $\left(0,\frac{2(2\sqrt{3}-3)}{3}\right),(2,\infty )\text{ and concave down on }\left(\frac{2(2\sqrt{3}-3)}{3},2\right)$.

Step 4. Verification.

The graph of the function is,