Chapter 0.5, Problem 35E

To determine

To solve: An equation algebrically and verify solution graphically.

Answer to Problem 35E

The required algebraic and graphical solutions are $x=\pm 0.962$ .

Explanation of Solution

Given data: The given equation is $e^x+e^{-x}=3$ .

Formula used:

The solution of a quadratic equation $a{y}^2+by+c=0$ is given by $y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Calculation:

The given equation is $e^x+e^{-x}=3$ or

  $e^x+\frac{1}{e^x}=3$ … (1)

Let $e^x=y$ , then equation (1) may be written as

  $\begin{array}{c}y+\frac{1}{y}=3\leftarrow \text{multiply}\text{both}\text{sides}\text{by}y\\ y^2+1=3y\\ y^2-3y+1=0\end{array}$

The comparing quadratic equation $y^2-3y+1=0$ with $a{y}^2+by+c=0$ , $a=1,b=-3,c=1$ . Therefore, the solution of equation $y^2-3y+1=0$ is,

  $\begin{array}{c}y=\frac{-\left(-3\right)\pm \sqrt{{\left(-3\right)}^2-4\left(1\right)1}}{2\cdot 1}\\ =\frac{3\pm \sqrt{9-4}}{2}\\ =\frac{3\pm \sqrt{5}}{2}\\ =2.6180,0.3820\end{array}$

If $y=2.6180$ , then $e^x=2.6180$ , taking log at the base e ,

  $\begin{array}{c}\ln \left(e^x\right)=\ln \left(2.6180\right)\\ x\ln \left(e\right)=0.9624\\ x=0.9624\because \ln \left(e\right)=1\end{array}$

If $y=0.3820$ , then $e^x=0.3820$ , taking log at the base e ,

  $\begin{array}{c}\ln \left(e^x\right)=\ln \left(0.3820\right)\\ x\ln \left(e\right)=-0.9623\\ x=-0.9623\because \ln \left(e\right)=1\end{array}$

Thus, the algebraic solutions of the equation $e^x+e^{-x}=3$ are $x=\pm 0.962$ .

Now, let $y\left(x\right)=e^x+e^{-x}-3$ . Then $y\left(x\right)=0$ is the solution of equation $e^x+e^{-x}=3$ . Therefore, the values of x for which $y\left(x\right)=0$ is the solution of $e^x+e^{-x}=3$ .

The graph of equation $e^x+e^{-x}=3$ is shown in Fig.1

  

The graph of function $y\left(x\right)=e^x+e^{-x}-3$ intersect x -axis at points $\left(0.962,0\right)$ and $\left(-0.962,0\right)$ . Therefore, $x=\pm 0.962$ are the graphical solutions of the equation $e^x+e^{-x}=3$ .

Thus, the algebraic and the graphical solutions of the equation $e^x+e^{-x}=3$ are identical given by $x=\pm 0.962$ .