Chapter 4.3, Problem 19E

To determine

To calculate : The points of inflection of the function $y=\frac{x^3-2x^2+x-1}{x-2}$ .

Answer to Problem 19E

The point of inflection of the function $y=\frac{x^3-2x^2+x-1}{x-2}$ is $\left(1,1\right)$ .

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, quotient rule to differentiate.

Calculation :

Consider the function $y=\frac{x^3-2x^2+x-1}{x-2}$ .

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

  $\begin{array}{c}{y}^{\prime }=\frac{\left(x-2\right)\frac{d}{dx}\left(x^3-2x^2+x-1\right)-\left(x^3-2x^2+x-1\right)\frac{d}{dx}\left(x-2\right)}{{\left(x-2\right)}^2}\\ =\frac{\left(x-2\right)\left(3x^2-4x+1\right)-\left(x^3-2x^2+x-1\right)\left(1\right)}{{\left(x-2\right)}^2}\\ =\frac{3x^3-4x^2+x-6x^2+8x-2-x^3+2x^2-x+1}{{\left(x-2\right)}^2}\\ =\frac{2x^3-8x^2+8x-1}{{\left(x-2\right)}^2}\end{array}$

Calculate the second derivative of the function using power rule.

  $\begin{array}{c}{y}^{″}=\frac{{\left(x-2\right)}^2\frac{d}{dx}\left(2x^3-8x^2+8x-1\right)-\left(2x^3-8x^2+8x-1\right)\frac{d}{dx}{\left(x-2\right)}^2}{{\left({\left(x-2\right)}^2\right)}^2}\\ =\frac{{\left(x-2\right)}^2\left(6x^2-16x+8\right)-\left(2x^3-8x^2+8x-1\right)\left(2\right)\left(x-2\right)}{{\left(x-2\right)}^4}\\ =\frac{\left(x-2\right)\left(6x^2-16x+8\right)-\left(2x^3-8x^2+8x-1\right)\left(2\right)}{{\left(x-2\right)}^3}\\ =\frac{2\left(x^3-6x^2+12x-7\right)}{{\left(x-2\right)}^3}\end{array}$

Set $y^{″}=0$ and solve the equation.

  $\begin{array}{c}\frac{2\left(x^3-6x^2+12x-7\right)}{{\left(x-2\right)}^3}=0\\ x^3-6x^2+12x-7=0\end{array}$

It can be seen that for the value $x=1$ , the value comes out to be 0.

  $\begin{array}{c}{1}^3-6{\left(1\right)}^2+12\left(1\right)-7=1-6+12-7\\ =13-13\\ =0\end{array}$

The inflection point is $\left(1,f\left(1\right)\right)$ .

Now,

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

Draw the graph of the function $y=\frac{x^3-2x^2+x-1}{x-2}$ .

  

It can also be seen from the graph that the concavity of the graph changes at the point $\left(1,1\right)$ .

Thus, the point of inflection of the function $y=\frac{x^3-2x^2+x-1}{x-2}$ is $\left(1,1\right)$ .