>Applications of the Derivative>Analyzing Exponential Functions

Analyzing Exponential Functions

Now we’ll take a look at an example where we analyze an exponential function. The method is as follows:

Rule

Analyzing Exponential Functions

1.: Find the zeros.
2.: Find the stationary points.
3.: Find the inflection points.

Example 1

Analyze the function $f (x) = 2 x^{2} \cdot e^{x}$

Example of analysis of exponential function

1.

Find the zeros by setting

f (x) = 0

2 x^{2} \cdot e^{x} = 0

The zero product property gives that $2 x^{2} = 0$ or $e^{x} = 0$ . However, $e^{x}$ is always positive, so

\begin{array}{l} 2 x^{2} & = 0 \\ x^{2} & = 0 \\ x & = 0 \end{array}

This gives a zero at the origin $(0, 0)$ .

2.

Find the maxima and minima by setting

f^{'} (x) = 0

. First, find the derivative of

f (x) = 2 x^{2} \cdot e^{x}

\begin{array}{l} f^{'} (x) & = 4 x \cdot e^{x} + 2 x^{2} \cdot e^{x} \\ = e^{x} (4 x + 2 x^{2}) \\ = 2 x \cdot e^{x} (2 + x) \end{array}

Then, find where $f^{'} (x)$ is equal to 0:

2 x \cdot e^{x} (2 + x) = 0

Again, $e^{x}$ is always positive, so

\begin{array}{l} 2 x & = 0 & \Rightarrow & x & = 0 \\ 2 + x & = 0 & \Rightarrow & x & = - 2 \end{array}

You then need the corresponding $y$ -values to find the point. You do this by inputting your $x$ -values back into the main function $f (x)$ :

\begin{array}{l} y & = f (0) = 2 \cdot 0^{2} \cdot e^{0} = 0 \\ y & = f (- 2) \\ = 2 {(- 2)}^{2} \cdot e^{- 2} = 8 e^{- 2} \\ = \frac{8}{e^{2}} \end{array}

\begin{array}{l} y & = f (0) = 2 \cdot 0^{2} \cdot e^{0} = 0 \\ y & = f (- 2) = 2 {(- 2)}^{2} \cdot e^{- 2} = 8 e^{- 2} = \frac{8}{e^{2}} \end{array}

You now need to determine which point is a maximum and which is a minimum. You do that by drawing a sign chart.

Sign chart of exponential function

From this, you see that the maximum is

(- 2, \frac{8}{e^{2}})

and the minimum is

(0, 0)

3.

Find the inflection points by setting

f^{″} (x) = 0

. First, you find the second derivative by differentiating

f^{'} (x) = e^{x} (4 x + 2 x^{2})

\begin{array}{l} f^{″} (x) & = e^{x} (4 x + 2 x^{2}) + e^{x} (4 + 4 x) \\ = e^{x} (4 x + 2 x^{2} + 4 + 4 x) \\ = e^{x} (2 x^{2} + 8 x + 4) \end{array}

\begin{array}{l} f^{″} (x) & = e^{x} (4 x + 2 x^{2}) + e^{x} (4 + 4 x) \\ = e^{x} (4 x + 2 x^{2} + 4 + 4 x) \\ = e^{x} (2 x^{2} + 8 x + 4) \end{array}

Then, let

f^{″} (x) = 0

and solve the equation:

e^{x} (2 x^{2} + 8 x + 4) = 0

As $e^{x}$ is always positive, you get

2 x^{2} + 8 x + 4 = 0

You solve this using the quadratic formula and get the solutions $x \approx - 0.6$ and $x \approx - 3.4$ . You find the corresponding $y$ -values by putting your new $x$ -values back into the main function $f (x)$ . You then get:

\begin{array}{l} y & = f (- 3.4) \\ = 2 \cdot {(- 3.4)}^{2} \cdot e^{- 3.4} \\ \approx 0.772 \\ y & = f (- 0.6) \\ = 2 {(- 0.6)}^{2} \cdot e^{- 0.6} \\ = 8 e^{- 0.6} \\ \approx 0.395 \end{array}

which means that you have inflection points at $(- 3.4, 0.772)$ and $(- 0.6, 0.395)$ .

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