Chapter 1.5, Problem 10QR

(a)

To determine

To find: The points of intersection of the two curves.

Answer to Problem 10QR

The intersection point is $\left(-1.39,4\right)$.

Explanation of Solution

Given information: The given equations are $y=e^{-x}$ and $y=4$.

Calculation:

Substitute $4$ for $y$ in the equation $y=e^{-x}$ to find the value of $x$.

  $4=e^{-x}$

Take $\ln$ on both sides of the equations.

  $\begin{array}{c}\ln \left(4\right)=\ln \left(e^{-x}\right)\\ \ln \left(4\right)=-x\ln \left(e\right)\\ -x=\ln \left(4\right)\\ x\approx -1.39\end{array}$

Substitute $-1.39$ for $x$ in the equation $y=e^{-x}$.

  $\begin{array}{c}y=e^{1.39}\\ =4\end{array}$

Therefore, the intersection point is $\left(-1.39,4\right)$.

(b)

To determine

To find: The points of intersection of the two curves.

Answer to Problem 10QR

There are no intersection points.

Explanation of Solution

Given information: The given equations are $y=e^{-x}$ and $y=-1$.

Calculation:

Since $e^{-x}>0$ for all real numbers $x$ , there are no solutions to the equation $-1=2^x$ . So, there are no intersection points.

Therefore, there are no intersection points.