Chapter 5.2, Problem 77E

To determine

Whether the function is even, odd, neither.

Answer to Problem 77E

The function $f\left(x\right)=\left|x\right|\cos x$ is an Even function.

Explanation of Solution

Given information:

  $f\left(x\right)=\left|x\right|\cos x$

Formula used:

  $\begin{array}{l}f\left(x\right)=f\left(-x\right):\text{ Condition for even function}\\ \\ f\left(x\right)=-f\left(x\right):\text{ Condition for odd function}\\ \\ \cos \left(-\theta \right)=\cos \theta \\ \\ \left|-x\right|=x\end{array}$

  $f\left(x\right)=\left|x\right|\cos x$

Now we know that if we put $x=-x$ in the equation and it results into $f\left(x\right)=f\left(-x\right)$ the equation is “Even”, and if the equation results into $f\left(x\right)=-f\left(x\right)$ the equation is odd.

Therefore, putting value of $x=-x$ in the equation to find the type of the equation:

  $\begin{array}{l}f\left(-x\right)=\left|-x\right|\cos \left(-x\right)\\ f(-x)=x\cos x\because \left|-x\right|=x,\cos \left(-\theta \right)=\cos \theta \\ \therefore f\left(-x\right)=f\left(x\right)\end{array}$

Hence, the function $f\left(x\right)=\left|x\right|\cos x$ is an Even function.