Chapter 8.1, Problem 9QR

To determine

To find the values of x at which the function changes sign on the given interval and to sketch the graph of the interval, and indicate the sign of the function on each subinterval.

Answer to Problem 9QR

The function changes sign at $\begin{array}{l}x=0.9633+k\pi \\ 2.1783+k\pi ,k\text{ is an intege}\end{array}$

Explanation of Solution

Given information:

The given function is $\sec \left(1+\sqrt{1-{\sin }^{2}x}\right)\text{ on }\left(-\infty ,\infty \right)$

Concept used:

Values at which a function changes its sign are the roots of that function.

Calculation :

To obtain the roots equate the function to 0.

  $\begin{array}{l}\sec \left(1+\sqrt{1-{\sin }^{2}x}\right)=0\\ \sec \left(1+\cos x\right)=0\left[\sqrt{1-{\sin }^{2}x}=\cos x\right]\\ \frac{1}{\cos \left(1+\cos x\right)}=0\end{array}$

The function will never be zero because the numerator is always non-zero. Since $\cos x$ is zero whenever x is of the form $\frac{\pi }{2}+kx$ where k is an integer. So the value of x be such that

  $\begin{array}{l}1+\cos x=\frac{\pi }{2}+kx\\ x=\arccos \left(\frac{\pi }{2}+kx-1\right)\\ x=0.9633+k\pi \\ =2.1783+k\pi ,k\text{ is an integer}\end{array}$

The function changes sign at $\begin{array}{l}x=0.9633+k\pi \\ 2.1783+k\pi ,k\text{ is an intege}\end{array}$