Chapter 12.3, Problem 7MRE

To determine

Find the zeroes of the function:

  $F(x)=\frac{x^3-4x}{x^2+1}$

Answer to Problem 7MRE

  $x=0\text{ or }x=2$

Explanation of Solution

Given:

  $F(x)=\frac{x^3-4x}{x^2+1}$

Calculation:

Zeroes of a function are the values at which the function’s value is 0.

  $\begin{array}{l}\frac{x^3-4x}{x^2+1}=0\\ \text{⇒}{x}^3-4x=\mathrm{0.........}[\text{multiply each side by (}{x}^2+1)]\\ \text{⇒}x(x^2-4)=0\\ \text{⇒}x=0\text{ or }(x^2-4)=\mathrm{0....}[ab=0\iff a=0orb=0]\\ \text{⇒}x=0\text{ or }{x}^2=4\\ \text{⇒}x=0\text{ or }x=\pm 2\\ But\text{ x=-2 does not satisfy the equation}\text{.}\\ \text{So, }\\ x=0\text{ or }x=2\end{array}$

So, $x=0\text{ or }x=2$ are the zeroes of the function.