Chapter 0.3, Problem 4QQ

a.

To determine

The domain of $f$

Answer to Problem 4QQ

The domain of $f\left(x\right)=e^{-x}-2$ is $x\in ℝ$ .

Explanation of Solution

Given:

Given the function is, $f\left(x\right)=e^{-x}-2$

Calculation:

Find the domain of given function $f\left(x\right)=e^{-x}-2$ .

The domain of a function refers to the inputs that can be passed into a function and

generates the outputs.

  $f\left(x\right)=e^{-x}-2$

The domain of the expression is all real numbers except where the expression is undefined.

In this case, there is no real number that makes the expression undefined.

Interval Notation:

  $\left(-\infty ,\infty \right)$

Set-Builder Notation:

  $\left\{x\mid x\in ℝ\right\}$

The domain of the function $f\left(x\right)=e^{-x}-2$ includes all real numbers.

  $x\in ℝ$

b.

To determine

The range of $f$

Answer to Problem 4QQ

The range of $f\left(x\right)=e^{-x}-2$ is $y\in \left(-2,\infty \right)$ .

Explanation of Solution

Given:

Given the function is, $f\left(x\right)=e^{-x}-2$

Calculation:

Find the range of the function $f\left(x\right)=e^{-x}-2$ .

Graphs with a horizontal asymptote never intersect when $f\left(x\right)$ represents a transformation of an exponential function.

  $\left(1\right)$ When $x$ approaches negative infinity, $y$ approaches positive infinity.

  $\left(2\right)$ When $x$ approaches positive infinity, the graph approaches the horizontal asymptote $y=-2$ . This is because $a^{-\infty }=0$ when $a>1$ .

So, by using $\left(1\right)$ and $\left(2\right)$ , the range of the function $f\left(x\right)=e^{-x}-2$ is:

  $y\in \left(-2,\infty \right)$

Another method:

The range is the set of all valid $y$ values. Use the graph to find the range.

  

Interval Notation:

  $\left(-2,\infty \right)$

Set-Builder Notation:

  $\left\{y\mid y>-2\right\}$

Thus, the range of the function is $y\in \left(-2,\infty \right)$ .

c.

To determine

The zeros of $f$

Answer to Problem 4QQ

The zeros of $f\left(x\right)=e^{-x}-2$ is approximately $-0.693$ .

Explanation of Solution

Given:

Given the function is, $f\left(x\right)=e^{-x}-2$

Calculation:

Find the zeros of the function $f\left(x\right)=e^{-x}-2$ .

The zeroes are the $x$ -values for which the function $f\left(x\right)$ is $0$ :

  $e^{-x}-2=0$

Add $2$ to both sides,

  $e^{-x}=2$

Take the natural logarithm of both sides of the equation to remove the variable from the exponent:

  $\ln \left(e^{-x}\right)=\ln \left(2\right)$

Expand $\ln \left(e^{-x}\right)$ by moving $-x$ outside the logarithm:

  $-x\ln \left(e\right)=\ln \left(2\right)$

The natural logarithm of $e$ is $1$ :

  $\begin{array}{c}-x\cdot 1=\ln \left(2\right)\\ -x=\ln \left(2\right)\\ x=-\ln \left(2\right)\\ x\approx -0.693\end{array}$

In decimal, the value of $x$ is approximately $-0.693$ .