Chapter 4.3, Problem 18E

To determine

To calculate : The points of inflection of the function $y=x^{\frac{1}{2}}\left(x+2\right)$ .

Answer to Problem 18E

The points of inflection of the function $y=x^{\frac{1}{2}}\left(x+2\right)$ are $\left(0,0\right)$ and $\left(\frac{2}{3},\frac{2^{\frac{7}{2}}}{3^{\frac{3}{2}}}\right)$ .

Explanation of Solution

Given information : The function $y=x^{\frac{1}{2}}\left(x+2\right)$ .

Formula used : The point of inflection of the graph is the point where the function has a tangent line an where the concavity of the graph changes.

Use the power, product rule to differentiate.

Calculation :

Consider the function $y=x^{\frac{1}{2}}\left(x+2\right)$ ,

Calculate the first derivative of the function using product and power rule.

  $\begin{array}{c}{y}^{\prime }=x^{\frac{1}{2}}\frac{d}{dx}\left(x+2\right)+\left(x+2\right)\frac{d}{dx}{x}^{\frac{1}{2}}\\ =x^{\frac{1}{2}}\left(1\right)+\left(x+2\right)\left(\frac{1}{2}{x}^{-\frac{1}{2}}\right)\\ =x^{\frac{1}{2}}+\frac{1}{2}{x}^{\frac{1}{2}}+\frac{1}{x^{\frac{1}{2}}}\\ =\frac{3}{2}{x}^{\frac{1}{2}}+\frac{1}{x^{\frac{1}{2}}}\end{array}$

Calculate the second derivative of the function using power rule.

  $\begin{array}{c}{y}^{″}=\frac{d}{dx}\left(\frac{3}{2}{x}^{\frac{1}{2}}\right)+\frac{d}{dx}\left(\frac{1}{x^{\frac{1}{2}}}\right)\\ =\frac{3}{2}\left(\frac{1}{2x^{\frac{1}{2}}}\right)+\left(-\frac{1}{2x^{\frac{3}{2}}}\right)\\ =\frac{3}{4x^{\frac{1}{2}}}-\frac{1}{2x^{\frac{3}{2}}}\end{array}$

Set $y^{″}=0$ and solve equation.

  $\begin{array}{c}\frac{3}{4x^{\frac{1}{2}}}-\frac{1}{2x^{\frac{3}{2}}}=0\\ \frac{3}{4x^{\frac{1}{2}}}=\frac{1}{2x^{\frac{3}{2}}}\\ \frac{x^{\frac{3}{2}}}{x^{\frac{1}{2}}}=\frac{4}{2\left(3\right)}\\ x=\frac{2}{3}\end{array}$

The inflection points are $\left(0,f\left(0\right)\right)$ and $\left(\frac{2}{3},f\left(\frac{2}{3}\right)\right)$ .

Now,

  $\begin{array}{c}f\left(0\right)=y\left(0\right)\\ ={\left(0\right)}^{\frac{1}{2}}\left(0+2\right)\\ =0\end{array}$

And,

  $\begin{array}{c}f\left(\frac{2}{3}\right)=y\left(\frac{2}{3}\right)\\ ={\left(\frac{2}{3}\right)}^{\frac{1}{2}}\left(\frac{2}{3}+2\right)\\ ={\left(\frac{2}{3}\right)}^{\frac{1}{2}}\left(\frac{8}{3}\right)\\ =\frac{2^{\frac{7}{2}}}{3^{\frac{3}{2}}}\end{array}$

Draw the graph of the function $y=x^{\frac{1}{2}}\left(x+2\right)$ .

  

It can also be seen from the graph that the concavity of the graph changes at the point $\left(0,0\right)$ and $\left(\frac{2}{3},\frac{2^{\frac{7}{2}}}{3^{\frac{3}{2}}}\right)$ .

Thus, the points of inflection of the function $y=x^{\frac{1}{2}}\left(x+2\right)$ are $\left(0,0\right)$ and $\left(\frac{2}{3},\frac{2^{\frac{7}{2}}}{3^{\frac{3}{2}}}\right)$ .