Paul's Online Notes
Home / Calculus I / Applications of Derivatives / Critical Points
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 4-2 : Critical Points

Critical points will show up throughout a majority of this chapter so we first need to define them and work a few examples before getting into the sections that actually use them.

#### Definition

We say that $$x = c$$ is a critical point of the function $$f\left( x \right)$$ if $$f\left( c \right)$$ exists and if either of the following are true.

$f'\left( c \right) = 0\hspace{0.5in}{\mbox{OR}}\hspace{0.5in}f'\left( c \right)\,\,\,{\mbox{doesn't exist}}$

Note that we require that $$f\left( c \right)$$ exists in order for $$x = c$$ to actually be a critical point. This is an important, and often overlooked, point. What this is really saying is that all critical points must be in the domain of the function. If a point is not in the domain of the function then it is not a critical point.

Note as well that, at this point, we only work with real numbers and so any complex numbers that might arise in finding critical points (and they will arise on occasion) will be ignored. There are portions of calculus that work a little differently when working with complex numbers and so in a first calculus class such as this we ignore complex numbers and only work with real numbers. Calculus with complex numbers is beyond the scope of this course and is usually taught in higher level mathematics courses.

The main point of this section is to work some examples finding critical points. So, let’s work some examples.

Example 1 Determine all the critical points for the function. $f\left( x \right) = 6{x^5} + 33{x^4} - 30{x^3} + 100$
Show Solution

We first need the derivative of the function in order to find the critical points and so let’s get that and notice that we’ll factor it as much as possible to make our life easier when we go to find the critical points.

\begin{align*}f'\left( x \right) & = 30{x^4} + 132{x^3} - 90{x^2}\\ & = 6{x^2}\left( {5{x^2} + 22x - 15} \right)\\ & = 6{x^2}\left( {5x - 3} \right)\left( {x + 5} \right)\end{align*}

Now, our derivative is a polynomial and so will exist everywhere. Therefore, the only critical points will be those values of $$x$$ which make the derivative zero. So, we must solve.

$6{x^2}\left( {5x - 3} \right)\left( {x + 5} \right) = 0$

Because this is the factored form of the derivative it’s pretty easy to identify the three critical points. They are,

$x = - 5,\,\,\,\,\,\,x = 0,\,\,\,\,\,\,x = \frac{3}{5}$

Polynomials are usually fairly simple functions to find critical points for provided the degree doesn’t get so large that we have trouble finding the roots of the derivative.

Most of the more “interesting” functions for finding critical points aren’t polynomials however. So let’s take a look at some functions that require a little more effort on our part.

Example 2 Determine all the critical points for the function. $g\left( t \right) = \sqrt[3]{{{t^2}}}\left( {2t - 1} \right)$
Show Solution

To find the derivative it’s probably easiest to do a little simplification before we actually differentiate. Let’s multiply the root through the parenthesis and simplify as much as possible. This will allow us to avoid using the product rule when taking the derivative.

$g\left( t \right) = {t^{\frac{2}{3}}}\left( {2t - 1} \right) = 2{t^{\frac{5}{3}}} - {t^{\frac{2}{3}}}$

Now differentiate.

$g'\left( t \right) = \frac{{10}}{3}{t^{\frac{2}{3}}} - \frac{2}{3}{t^{ - \frac{1}{3}}} = \frac{{10{t^{\frac{2}{3}}}}}{3} - \frac{2}{{3{t^{\frac{1}{3}}}}}$

We will need to be careful with this problem. When faced with a negative exponent it is often best to eliminate the minus sign in the exponent as we did above. This isn’t really required but it can make our life easier on occasion if we do that.

Notice as well that eliminating the negative exponent in the second term allows us to correctly identify why $$t = 0$$ is a critical point for this function. Once we move the second term to the denominator we can clearly see that the derivative doesn’t exist at $$t = 0$$ and so this will be a critical point. If you don’t get rid of the negative exponent in the second term many people will incorrectly state that $$t = 0$$ is a critical point because the derivative is zero at $$t = 0$$. While this may seem like a silly point, after all in each case $$t = 0$$ is identified as a critical point, it is sometimes important to know why a point is a critical point. In fact, in a couple of sections we’ll see a fact that only works for critical points in which the derivative is zero.

So, we’ve found one critical point (where the derivative doesn’t exist), but we now need to determine where the derivative is zero (provided it is of course…). To help with this it’s usually best to combine the two terms into a single rational expression. So, getting a common denominator and combining gives us,

$g'\left( t \right) = \frac{{10t - 2}}{{3{t^{\frac{1}{3}}}}}$

Notice that we still have $$t = 0$$ as a critical point. Doing this kind of combining should never lose critical points, it’s only being done to help us find them. As we can see it’s now become much easier to quickly determine where the derivative will be zero. Recall that a rational expression will only be zero if its numerator is zero (and provided the denominator isn’t also zero at that point of course).

So, in this case we can see that the numerator will be zero if $$t = \frac{1}{5}$$ and so there are two critical points for this function.

$t = 0\hspace{0.5in}\,{\mbox{and}}\hspace{0.5in}t = \frac{1}{5}$
Example 3 Determine all the critical points for the function. $R\left( w \right) = \frac{{{w^2} + 1}}{{{w^2} - w - 6}}$
Show Solution

We’ll leave it to you to verify that using the quotient rule, along with some simplification, we get that the derivative is,

$R'\left( w \right) = \frac{{ - {w^2} - 14w + 1}}{{{{\left( {{w^2} - w - 6} \right)}^2}}} = - \frac{{{w^2} + 14w - 1}}{{{{\left( {{w^2} - w - 6} \right)}^2}}}$

Notice that we factored a “-1” out of the numerator to help a little with finding the critical points. This negative out in front will not affect the derivative whether or not the derivative is zero or not exist but will make our work a little easier.

Now, we have two issues to deal with. First the derivative will not exist if there is division by zero in the denominator. So we need to solve,

${w^2} - w - 6 = \left( {w - 3} \right)\left( {w + 2} \right) = 0$

We didn’t bother squaring this since if this is zero, then zero squared is still zero and if it isn’t zero then squaring it won’t make it zero.

So, we can see from this that the derivative will not exist at $$w = 3$$ and $$w = - 2$$. However, these are NOT critical points since the function will also not exist at these points. Recall that in order for a point to be a critical point the function must actually exist at that point.

At this point we need to be careful. The numerator doesn’t factor, but that doesn’t mean that there aren’t any critical points where the derivative is zero. We can use the quadratic formula on the numerator to determine if the fraction as a whole is ever zero.

$w = \frac{{ - 14 \pm \sqrt {{{\left( {14} \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{{2\left( 1 \right)}} = \frac{{ - 14 \pm \sqrt {200} }}{2} = \frac{{ - 14 \pm 10\sqrt 2 }}{2} = - 7 \pm 5\sqrt 2$

So, we get two critical points. Also, these are not “nice” integers or fractions. This will happen on occasion. Don’t get too locked into answers always being “nice”. Often they aren’t.

Note as well that we only use real numbers for critical points. So, if upon solving the quadratic in the numerator, we had gotten complex number these would not have been considered critical points.

Summarizing, we have two critical points. They are,

$w = - 7 + 5\sqrt 2 ,\,\,\,\,w = - 7 - 5\sqrt 2$

Again, remember that while the derivative doesn’t exist at $$w = 3$$ and $$w = - 2$$ neither does the function and so these two points are not critical points for this function.

In the previous example we had to use the quadratic formula to determine some potential critical points. We know that sometimes we will get complex numbers out of the quadratic formula. Just remember that, as mentioned at the start of this section, when that happens we will ignore the complex numbers that arise.

So far all the examples have not had any trig functions, exponential functions, etc. in them. We shouldn’t expect that to always be the case. So, let’s take a look at some examples that don’t just involve powers of $$x$$.

Example 4 Determine all the critical points for the function. $y = 6x - 4\cos \left( {3x} \right)$
Show Solution

First get the derivative and don’t forget to use the chain rule on the second term.

$y' = 6 + 12\sin \left( {3x} \right)$

Now, this will exist everywhere and so there won’t be any critical points for which the derivative doesn’t exist. The only critical points will come from points that make the derivative zero. We will need to solve,

\begin{align*}6 + 12\sin \left( {3x} \right) & = 0\\ \sin \left( {3x} \right) & = - \frac{1}{2}\end{align*}

Solving this equation gives the following.

\begin{align*}3x & = 3.6652 + 2\pi n,\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots \\ 3x & = 5.7596 + 2\pi n,\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots \end{align*}

Don’t forget the $$2 \pi n$$ on these! There will be problems down the road in which we will miss solutions without this! Also make sure that it gets put on at this stage! Now divide by 3 to get all the critical points for this function.

\begin{align*}x &= 1.2217 + \frac{{2\pi n}}{3},\hspace{0.5in}n = 0, \pm 1, \pm 2, \ldots \\ x &= 1.9199 + \frac{{2\pi n}}{3},\hspace{0.5in}n = 0, \pm 1, \pm 2, \ldots \end{align*}

Notice that in the previous example we got an infinite number of critical points. That will happen on occasion so don’t worry about it when it happens.

Example 5 Determine all the critical points for the function. $h\left( t \right) = 10t{{\bf{e}}^{3 - {t^2}}}$
Show Solution

Here’s the derivative for this function.

$h'\left( t \right) = 10{{\bf{e}}^{3 - {t^2}}} + 10t{{\bf{e}}^{3 - {t^2}}}\left( { - 2t} \right) = 10{{\bf{e}}^{3 - {t^2}}} - 20{t^2}{{\bf{e}}^{3 - {t^2}}}$

Now, this looks unpleasant, however with a little factoring we can clean things up a little as follows,

$h'\left( t \right) = 10{{\bf{e}}^{3 - {t^2}}}\left( {1 - 2{t^2}} \right)$

This function will exist everywhere, so no critical points will come from the derivative not existing. Determining where this is zero is easier than it looks. We know that exponentials are never zero and so the only way the derivative will be zero is if,

\begin{align*}1 - 2{t^2} & = 0\\ 1 & = 2{t^2}\\ \frac{1}{2} & = {t^2}\end{align*}

We will have two critical points for this function.

$t = \pm \frac{1}{{\sqrt 2 }}$
Example 6 Determine all the critical points for the function. $f\left( x \right) = {x^2}\ln \left( {3x} \right) + 6$
Show Solution

Before getting the derivative let’s notice that since we can’t take the log of a negative number or zero we will only be able to look at $$x > 0$$.

The derivative is then,

\begin{align*}f'\left( x \right) & = 2x\ln \left( {3x} \right) + {x^2}\left( {\frac{3}{{3x}}} \right)\\ & = 2x\ln \left( {3x} \right) + x\\ & = x\left( {2\ln \left( {3x} \right) + 1} \right)\end{align*}

Now, this derivative will not exist if $$x$$ is a negative number or if $$x = 0$$, but then again neither will the function and so these are not critical points. Remember that the function will only exist if $$x > 0$$ and nicely enough the derivative will also only exist if $$x > 0$$ and so the only thing we need to worry about is where the derivative is zero.

First note that, despite appearances, the derivative will not be zero for $$x = 0$$. As noted above the derivative doesn’t exist at $$x = 0$$ because of the natural logarithm and so the derivative can’t be zero there!

So, the derivative will only be zero if,

\begin{align*}2\ln \left( {3x} \right) + 1 & = 0\\ \ln \left( {3x} \right) & = - \frac{1}{2}\end{align*}

Recall that we can solve this by exponentiating both sides.

\begin{align*}{{\bf{e}}^{\ln \left( {3x} \right)}} & = {{\bf{e}}^{ - \frac{1}{2}}}\\ 3x & = {{\bf{e}}^{ - \frac{1}{2}}}\\ x & = \frac{1}{3}{{\bf{e}}^{ - \frac{1}{2}}} = \frac{1}{{3\sqrt {\bf{e}} }}\end{align*}

There is a single critical point for this function.

Let’s work one more problem to make a point.

Example 7 Determine all the critical points for the function. $f\left( x \right) = x{{\bf{e}}^{{x^2}}}$
Show Solution

Note that this function is not much different from the function used in Example 5. In this case the derivative is,

$f'\left( x \right) = {{\bf{e}}^{{x^2}}} + x{{\bf{e}}^{{x^2}}}\left( {2x} \right) = {{\bf{e}}^{{x^2}}}\left( {1 + 2{x^2}} \right)$

This function will never be zero for any real value of $$x$$. The exponential is never zero of course and the polynomial will only be zero if $$x$$ is complex and recall that we only want real values of $$x$$ for critical points.

Therefore, this function will not have any critical points.

It is important to note that not all functions will have critical points! In this course most of the functions that we will be looking at do have critical points. That is only because those problems make for more interesting examples. Do not let this fact lead you to always expect that a function will have critical points. Sometimes they don’t as this final example has shown.