Chapter 3.1, Problem 48E

Solution

To graph: The given function and to analyze it.

  

The domain is $\left(-\infty ,\infty \right)$ .

The range is $\left(0,\infty \right)$ .

It is a continuous function for all real numbers.

The function is always increasing.

No symmetry.

Bounded below, but not above.

No local extrema.

Horizontal asymptote is $y=0$ .

No vertical asymptote.

End behavior: $\underset{x\to -\infty }{\lim }f\left(x\right)=\infty ,\underset{x\to \infty }{\lim }f\left(x\right)=0$ .

Given:

  $f\left(x\right)=5\cdot e^{-x}$

Calculation:

Comparing $f\left(x\right)=5\cdot e^{-x}$ with $f\left(x\right)=a\cdot e^{kx}$ , it follows that

  $a=5,k=-1$ .

Since $a>0$ and $k<0$ , therefore it is an exponential decay function.

Now, the graph is

  

The given function is well defined for all real numbers, therefore the domain is $\left(-\infty ,\infty \right)$ .

From the graph, it can be seen that the $y$ -values are ranging from 0 to $\infty$ .

Therefore, the range is $\left(0,\infty \right)$ .

From the graph, it can be seen that the curve has no breaks or jumps; therefore it is a continuous function for all real numbers.

From the graph, it can be seen that the value of $y$ decreases as the value of $x$ increases from $-\infty$ to $\infty$ .So, the function is always decreasing.

The curve is neither symmetric with the $x$ -axis nor with the $y$ -axis.

So the curve has no symmetry.

Bounded below, but not above.

There are no turning points for the given curve, so there is no local extrema.

Horizontal asymptote is $y=0$ .

No vertical asymptote.

End behavior:

From the graph, it can be seen that $y\to \infty$ as $x\to -\infty$ and $y\to 0$ as $x\to \infty$ . So,

  $\begin{array}{c}\underset{x\to -\infty }{\lim }f\left(x\right)=\infty \\ \underset{x\to \infty }{\lim }f\left(x\right)=0\end{array}$