Tangents and normals to exponential functions
We have just explored the differentiation of exponential functions. As with the derivatives of other functions which we have seen, we can use this to find the equation of tangents and normals to exponential functions.
| $\frac{d}{dx}e^x$ddxex | $=$= | $e^x$ex |
| $\frac{d}{dx}e^{ax}$ddxeax | $=$= | $ae^{ax}$aeax |
| $\frac{d}{dx}e^{f\left(x\right)}$ddxef(x) | $=$= | $f'\left(x\right)e^{f\left(x\right)}$f′(x)ef(x) |
When finding equations of tangents and normals, remember that the gradient of the tangent at any point on a function is given by the first derivative of the function. The following formulas are also useful:
Given the gradient $m$m and a point $\left(x_1,y_1\right)$(x1,y1), the equation of a line can be found using the point gradient formula:
$y-y_1=m\left(x-x_1\right)$y−y1=m(x−x1)
Given the gradient of the tangent $m_1$m1 at a point, the gradient of the normal $m_2$m2 is:
$m_2=\frac{-1}{m_1}$m2=−1m1
Worked examples
Example 1
Find the equation of the tangent and the normal to the curve given by $y=e^{x^2-4x}$y=ex2−4x at the point where $x=4$x=4.
Think: To find the equation of the tangent, we will use the point gradient formula. For this we need to find the $y$y coordinate on the function for $x=4$x=4.
Do:
| $y$y | $=$= | $e^{x^2-4x}$ex2−4x |
| $y$y | $=$= | $e^{4^2-16}$e42−16 |
| $y$y | $=$= | $e^0$e0 |
| $y$y | $=$= | $1$1 |
Next, we will determine the gradient function and evaluate the gradient at $x=4$x=4. We do this by finding the first derivative of the function as follows:
| $y$y | $=$= | $e^{x^2-4x}$ex2−4x |
| $\frac{dy}{dx}$dydx | $=$= | $(2x-4)e^{x^2-4x}$(2x−4)ex2−4x |
For $x=4$x=4:
| $\frac{dy}{dx}$dydx | $=$= | $(2\times4-4)e^{4^2-16}$(2×4−4)e42−16 |
| $\frac{dy}{dx}$dydx | $=$= | $4$4 |
Using the point gradient formula, with the point $\left(4,1\right)$(4,1) and gradient of $4$4 we can find the equation of the tangent as follows:
| $y-y_1$y−y1 | $=$= | $m\left(x-x_1\right)$m(x−x1) |
| $y-1$y−1 | $=$= | $4(x-4)$4(x−4) |
| $y-1$y−1 | $=$= | $4x-16$4x−16 |
| $y$y | $=$= | $4x-15$4x−15 |
To find the equation of the normal at the same point we first find the gradient. We know the gradient of the tangent $m_1=4$m1=4, hence the gradient of the normal $m_2$m2 is given by:
| $m_2$m2 | $=$= | $-\frac{1}{m_1}$−1m1 |
| $m_2$m2 | $=$= | $-\frac{1}{4}$−14 |
Using the point gradient formula, with the point $\left(4,1\right)$(4,1) and gradient of $-\frac{1}{4}$−14 we can find the equation of the tangent as follows:
| $y-y_1$y−y1 | $=$= | $m\left(x-x_1\right)$m(x−x1) |
| $y-1$y−1 | $=$= | $-\frac{1}{4}(x-4)$−14(x−4) |
| $y-1$y−1 | $=$= | $-\frac{x}{4}+1$−x4+1 |
| $y$y | $=$= | $-\frac{x}{4}+2$−x4+2 |
Practice question
Question 1
Find the equation of the tangent to the curve $f\left(x\right)=2e^x$f(x)=2ex at the point where it crosses the $y$y-axis.
Express the equation in the form $y=mx+c$y=mx+c.
Determining the characteristics of, and sketching exponential functions
The first and second derivatives of exponential functions can be used to determine concavity, and regions where an exponential function is increasing or decreasing. Additionally, any stationary points and points of inflection can be found. Using these tools we can sketch functions involving exponentials.
Worked example
Example 2
Sketch the function $f\left(x\right)=\frac{1}{2}e^{-x^2}$f(x)=12e−x2 showing any intercepts and stationary points.
Think: This function involves an even index and therefore it would be worth checking if it is an even function. If it is an even function we will be able to exploit the symmetrical nature.
Do: To check if it is an even function we test if $f\left(x\right)=f\left(-x\right)$f(x)=f(−x):
| $f\left(x\right)$f(x) | $=$= | $\frac{1}{2}e^{-x^2}$12e−x2 |
| $f\left(-x\right)$f(−x) | $=$= | $\frac{1}{2}e^{-(-x)^2}$12e−(−x)2 |
| $f\left(-x\right)$f(−x); | $=$= | $\frac{1}{2}e^{-x^2}$12e−x2 |
| $f\left(-x\right)$f(−x) | $=$= | $f\left(x\right)$f(x) |
Therefore, it is an even function and symmetrical about the $y$y-axis.
To find the $y$y-intercept, we make $x=0$x=0:
| $f\left(0\right)$f(0) | $=$= | $\frac{1}{2}e^{-0^2}$12e−02 |
| $f\left(0\right)$f(0) | $=$= | $\frac{1}{2}$12 |
To find any $x$x-intercepts, we make $y=0$y=0:
| $0$0 | $=$= | $\frac{1}{2}e^{-x^2}$12e−x2 |
There are no solutions to this equation and hence, there are no $x$x-intercepts.
To check for any stationary points we need to find the first derivative and make it equal to $0$0:
| $f\left(x\right)$f(x) | $=$= | $\frac{1}{2}e^{-x^2}$12e−x2 |
| $f'\left(x\right)$f′(x) | $=$= | $-2x\frac{1}{2}e^{-x^2}$−2x12e−x2 |
| $f'\left(x\right)$f′(x) | $=$= | $-xe^{-x^2}$−xe−x2 |
| $0$0 | $=$= | $-xe^{-x^2}$−xe−x2 |
| $x$x | $=$= | $0$0 |
Therefore, there is a stationary point at $x=0$x=0.
We will find the second derivative and use it to determine the nature of the stationary point and check for any points of inflection:
| $f''\left(x\right)$f′′(x) | $=$= | $-e^{-x^2}+-2xe^{-x^2}(-x)$−e−x2+−2xe−x2(−x) |
| $f''\left(x\right)$f′′(x) | $=$= | $-e^{-x^2}+2x^2e^{-x^2}$−e−x2+2x2e−x2 |
| $f''\left(x\right)$f′′(x) | $=$= | $e^{-x^2}\left(2x^2-1\right)$e−x2(2x2−1) |
At $x=0$x=0:
| $f''\left(0\right)$f′′(0) | $=$= | $e^{-2(0)^2}\left(2(0)^2-1\right)$e−2(0)2(2(0)2−1) |
| $f''\left(0\right)$f′′(0) | $=$= | $-1$−1 |
Therefore, there is a local maximum at $x=0$x=0 because $f''\left(x\right)<0$f′′(x)<0 and hence the curve is concave down at this point.
Setting $f''\left(x\right)=0$f′′(x)=0, we we can check for any inflection points:
| $0$0 | $=$= | $e^{-x^2}\left(2x^2-1\right)$e−x2(2x2−1) |
| $0$0 | $=$= | $2x^2-1$2x2−1 |
| $2x^2$2x2 | $=$= | $1$1 |
| $x^2$x2 | $=$= | $\frac{1}{2}$12 |
| $x$x | $=$= | $\pm\frac{1}{\sqrt{2}}$±1√2 |
Therefore, there are possible points of inflection at $x=\pm\frac{1}{\sqrt{2}}$x=±1√2.
We could check either the first derivative for the sign of the gradient or the second derivative for a change in concavity, to the left and right of these values, to determine if they are points of inflection. We will look at both below, but in practice, you would choose the simplest option.
| x | $-1$−1 | $-\frac{1}{\sqrt{2}}$−1√2 | $0$0 | $\frac{1}{\sqrt{2}}$1√2 | $1$1 |
|---|---|---|---|---|---|
| $f'\left(x\right)$f′(x) | $0.37$0.37 | $0.43$0.43 | $0$0 | $-0.43$−0.43 | $-0.37$−0.37 |
| $f''\left(x\right)$f′′(x) | $0.37$0.37 | $0$0 | $-1$−1 | $0$0 | $0.37$0.37 |
As the gradient remains the same and there is a change in concavity at both $x=\pm\frac{1}{\sqrt{2}}$x=±1√2 we can classify the points as points of inflection.
Finally, we will consider the behaviour of the function as $x\rightarrow\pm\infty$x→±∞, first, we will rearrange $f\left(x\right)$f(x) to make the function easier to analyse:
| $f\left(x\right)$f(x) | $=$= | $\frac{1}{2}e^{-x^2}$12e−x2 |
| $f\left(x\right)$f(x) | $=$= | $\frac{1}{2e^{x^2}}$12ex2 |
As $x\rightarrow\infty$x→∞, the denominator of the function will become very large and positive, hence the function will approach zero from above:
| As $x\rightarrow\infty$x→∞ | $f\left(x\right)\rightarrow0$f(x)→0 |
We could use symmetry and say the behaviour will be the same as $x\rightarrow-\infty$x→−∞. Let's consider $x\rightarrow-\infty$x→−∞ anyway. The denominator of the function will become very large and positive, hence the function will approach zero from above:
| As $x\rightarrow-\infty$x→−∞ | $f\left(x\right)\rightarrow0$f(x)→0 |
All that is left is to sketch the function showing all the important points:
Practice questions
Question 2
Consider the function $f\left(x\right)=3-e^{-x}$f(x)=3−e−x.
Determine $f\left(0\right)$f(0).
Determine $f'\left(0\right)$f′(0).
Which of the following statements is true?
$f'\left(x\right)<0$f′(x)<0 for $x\ge0$x≥0
A$f'\left(x\right)<0$f′(x)<0 for all real $x$x.
B$f'\left(x\right)>0$f′(x)>0 for all real $x$x.
C$f\left(x\right)>0$f(x)>0 for all real $x$x.
DDetermine the value of $\lim_{x\to\infty}f\left(x\right)$limx→∞f(x).
Question 3
Consider the function $f\left(x\right)=4e^{-x^2}$f(x)=4e−x2.
Determine $f'\left(x\right)$f′(x).
Determine the values of $x$x for which $f'\left(x\right)=0$f′(x)=0.
Determine the values of $x$x for which $f'\left(x\right)>0$f′(x)>0.
Determine the values of $x$x for which $f'\left(x\right)<0$f′(x)<0.
Determine the value of $\lim_{x\to\infty}f\left(x\right)$limx→∞f(x).
Determine the value of $\lim_{x\to-\infty}f\left(x\right)$limx→−∞f(x).
Which of the following is the graph of $f\left(x\right)$f(x)?
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