Chapter 4.6, Problem 62E

To determine

To sketch: the graph of the function, to find whether the function is even, odd or neither.

Answer to Problem 62E

The function $f\left(x\right)=x^2-\sec x$ is even.

Explanation of Solution

Given information:

Given function

  $f\left(x\right)=x^2-\sec x$

Calculation:

Let us consider the points to find whether the function $f\left(x\right)=x^2-\sec x$ is even, odd, or neither.

Let us, first plot the graph for $f\left(x\right)=x^2-\sec x$ .

  

From the above graph, we can see that the curve of the function $x^2-\sec x$ is symmetrical about the $y$ axis.

Thus, the function $x^2-\sec x$ is even.

Now we will verify algebraically whether the function $f\left(x\right)=x^2-\sec x$ is even, odd or neither.

We know, the function is even is $f\left(-x\right)\ne -f\left(x\right)$

Let us take $x=0$

Substituting $x=0$ in $f\left(-x\right)$

We get,

  $\begin{array}{c}f\left(-x\right)=-x^2-\sec \left(-x\right)\\ f\left(0\right)=-0-\sec \left(-0\right)\\ =-1\end{array}$

Also, substituting $x=0$ in $-f\left(x\right)$

We get,

  $\begin{array}{c}-f\left(x\right)=-\left(x^2-\sec x\right)\\ -f\left(0\right)=-\left(0-\sec 0\right)\\ =-\left(-1\right)\\ =1\end{array}$

Since, $f\left(-x\right)\ne -f\left(x\right)$

Hence the function $f\left(x\right)=x^2-\sec x$ is even.