Introduction

Many of the applications that we will explore in this chapter require us to identify the critical points of a function. In this section, we will define what a critical point is, and practice finding the critical points of various functions, both algebraically and graphically.

To motivate the definition, let's recall some of the rate of change examples we have already done. In order to discover the intervals over which a function increases or decreases, we first needed to find the points where the derivative was zero. These are examples of critical points.

Examples

Polynomials are usually fairly simple functions to find the critical points for, provided the degree isn't too large. Most of the more "interesting" functions require a little more effort to find their critical numbers. Let's look at a few.

Before we move on to the next example, let's take a look at the graph of the function in Example 2. This function has one critical number (at $t=0$) at which the derivative is undefined, and one critical number ($t=\frac15$) at which the derivative equals zero.

On this graph, we see the cusp at the origin, which indicates that the function is not differentiable at $t=0$. We then see that at the second critical point, the graph has a local minimum. We can also see that the derivative is zero at this point because the tangent line is horizontal there.

In the last example, we had to use the Quadratic Formula to determine some of the potential critical numbers. When we use the Quadratic Formula to solve a quadratic equation it also happens sometimes that the solutions are complex numbers. Since for us a critical number has to be a real number, we would disregard any complex solutions that we find.

The next few examples include trigonometric, exponential and logarithmic functions. Many of the applications we examine later in the chapter will involve these types of functions.

In the last example we found an infinite number of critical numbers. This happens on occasion, especially when working with trig functions.

We'll work one more example, just to make a point.

This last example illustrates the fact that not all functions have critical points. Most of the functions we will work with will have critical points, though, since those will make the more interesting examples.

Practice Problems

Determine all critical numbers of each of the following functions.

1. $f(x)=8x^3+81x^2-42x-8$

   Click for solution

   First we need the derivative, and it's helpful if it is in factored form. So, $$f'(x) = 24x^2 +162x-42 = 6(4x^2+27x-7) = 6(x+7)(4x-1)$$

   Recall that critical numbers are the values of the variable where the derivative is zero or does not exist. In this problem, the derivative is a polynomial, so we know that it exists everywhere. All we need to do then is set the derivative equal to zero and solve: $$6(x+7)(4x-1)=0 \hspace{.25in}\Rightarrow \hspace{.25in} \require{bbox} \bbox[2pt,border:2px solid green]{x=-7, x=\frac14}$$

2. $R(t)=1+80t^3+5t^4-2t^5$

   Click for solution

   First we need the derivative, and it's helpful if it is in factored form. So,

   \begin{align} R'(t) &= 240t^2+20t^3-10t^4\\ &= -10t^2(-24-2t+t^2)\\ &= -10t^2(t+4)(t-6)\\ \end{align}

   Recall that critical numbers are the values of the variable where the derivative is zero or does not exist. In this problem, the derivative is a polynomial, so we know that it exists everywhere. All we need to do then is set the derivative equal to zero and solve: $$-10t^2(t+4)(t-6)=0 \hspace{.25in}\Rightarrow \hspace{.25in} \require{bbox} \bbox[2pt,border:2px solid green] {t=0, t=-4, t=6}$$

3. $g(w)=2w^3-7w^2-3w-2$

   Click for solution

   First we need the derivative: $$g'(w) = 6w^2-14w-3$$ This time, we have a polynomial that doesn't factor easily, so let's leave it in this form.

   Recall that critical numbers are the values of the variable where the derivative is zero or does not exist. In this problem, the derivative is a polynomial, so we know that it exists everywhere. All we need to do then is set the derivative equal to zero and solve: $$6w^2-14w-3=0 \hspace{.25in}\Rightarrow \hspace{.25in} w=\frac{14\pm\sqrt{268}}{12}=\require{bbox} \bbox[2pt,border:2px solid green]{\frac{7\pm\sqrt{67}}{6}}$$

   As we can see, we used the quadratic formula to find the zeros of the derivative. Not all quadratics will factor over the integers, so don't forget that you can use the quadratic formula!

4. $g(x)=x^6-2x^5+8x^4$

   Click for solution

   First we need the derivative, and it's helpful if it is in factored form. So, $$g'(x) = 6x^5-10x^4+32x^3 = 3x^3(3x^2-5x+16)$$

   Recall that critical numbers are the values of the variable where the derivative is zero or does not exist. In this problem, the derivative is a polynomial, so we know that it exists everywhere. All we need to do then is set the derivative equal to zero and solve: $$3x^3(3x^2-5x+16)=0 \hspace{.25in}\Rightarrow \hspace{.25in} 3x^3=0 \hspace{.15in}\text{or}\hspace{.15in} 3x^2-5x+16=0$$

   The first of these equations tells us that $x=0$ is a critical number. To solve the second equation, we need the quadratic formula again. $$x=\frac{5\pm\sqrt{-167}}{6} = \frac{5\pm\sqrt{167}i}{6}$$

   Recall that critical numbers for us are only real numbers, not complex numbers. So the solutions of the second equation are not critical numbers. Therefore, the only critical number is

   $$\require{bbox} \bbox[2pt,border:2px solid green]{x = 0}$$

5. $h(z)=4z^3-3z^2+9z+12$

   Click for solution

   First we need the derivative: $$h'(z) = 12z^2-6z+9 = 3(4z^2-2z+3)$$

   Recall that critical numbers are the values of the variable where the derivative is zero or does not exist. In this problem, the derivative is a polynomial, so we know that it exists everywhere. All we need to do then is set the derivative equal to zero and solve: $$4z^2-2z+3=0 \hspace{.25in}\Rightarrow \hspace{.25in} z=\frac{2\pm\sqrt{-44}}{8}=\frac{1\pm\sqrt{11}i}{4}$$

   Recall that critical numbers for us are only real numbers, not complex numbers. Therefore this function has no critical numbers.

   Don't worry about there being no critical numbers for this function. There is no rule that says that every function has to have critical points!

6. $Q(x)=(2-8x)^4(x^2-9)^3$

   Click for solution

   First we need to find the derivative. We'll use the product rule and chain rule:

   \begin{align} Q'(x) &= 4(2-8x)^3(-8)(x^2-9)^3 + (2-8x)^4(3)(x^2-9)^2(2x)\\ &= 2(2-8x)^3(x^2-9)^2[-16(x^2-9)+3x(2-8x)]\\ &= 2(2-8x)^3(x^2-9)^2[-40x^2+6x+144]\\ &= -4(2-8x)^3(x^2-9)^2[20x^2-3x-72]\\ \end{align}

   Factoring the derivative as much as possible helps with the next step, and on this problem, that extra factoring is all but required to make the problem possible to finish.

   Recall that critical numbers are the values of the variable where the derivative is zero or does not exist. In this problem, the derivative is a polynomial, so we know that it exists everywhere. All we need to do then is set the derivative equal to zero:

   $$-4(2-8x)^3(x^2-9)^2[20x^2-3x-72]=0$$

   This leads to the following three equations that we need to solve:

   \begin{align} (2-8x)^3 &= 0\\ (x^2-9)^2 &= 0\\ 20x^2-3x-72 &= 0\\ \end{align}

   To solve the first two of these equations, we set the quantity inside the parentheses equal to zero since the exponent won't affect the solution. To solve the third equation, we use the quadratic formula.

   \begin{align} 2-8x=0 &\Rightarrow x=\frac14\\ x^2-9=0 &\Rightarrow x=\pm 3\\ 20x^2-3x-72=0 &\Rightarrow x=\frac{3\pm\sqrt{9-4(20)(-72)}}{40}\\ x &= \frac{3\pm\sqrt{5769}}{40}\\ \end{align}

   So we have 5 critical numbers:

   $$\require{bbox} \bbox[2pt,border:2px solid green]{x = \frac14, x=\pm 3, x=\frac{3\pm\sqrt{5769}}{40}}$$

7. $f(z)=\displaystyle\frac{z+4}{2z^2+z+8}$

   Click for solution

   First we need to find the derivative. We'll use the quotient rule:

   \begin{align} f'(z) &= \frac{(1)(2z^2+z+8)-(z+4)(4z+1)}{(2z^2+z+8)^2}\\ &= \frac{-2z^2-16z+4}{(2z^2+z+8)^2}\\ &= \frac{-2(z^2+8z-2)}{(2z^2+z+8)^2}\\ \end{align}

   Recall that critical numbers are the values of the variable where the derivative is zero or does not exist. Since the derivative is a rational function, we know that the derivative will not exist at any value of $z$ that causes the denominator to be zero, and the derivative will equal zero at any value of $z$ in the domain that makes the numerator equal zero.

   So we need to set both the numerator and the denominator equal to zero and solve for $z$. Note that the "-2" that we factored out of the numerator will not affect where the numerator is zero. Also, the exponent on the expression in the denominator does not affect where the denominator is zero. So we have the following two equations:

   \begin{align} z^2+8z-2=0 &\Rightarrow z=\frac{-8\pm\sqrt{72}}{2} = -4\pm 3\sqrt{2}\\ 2z^2 +z+8=0 &\Rightarrow z=\frac{-1\pm\sqrt{-63}}{4} = \frac{-1\pm 3\sqrt{7}i}{4}\\ \end{align}

   Recall that critical numbers for us are only real numbers, not complex numbers. So the second equation doesn't contribute any critical numbers. That is, the only critical numbers of this function are the zeros of the derivative.

   Therefore, the critical numbers are

   $$\require{bbox} \bbox[2pt,border:2px solid green]{z= -4\pm 3\sqrt{2}}$$

8. $R(x)=\displaystyle\frac{1-x}{x^2+2x-15}$

   Click for solution

   First we need to find the derivative. We'll use the quotient rule:

   $$R'(x) = \frac{(-1)(x^2+2x-15)-(1-x)(2x+2)}{(x^2+2x-15)^2} = \frac{x^2-2x+13}{(x^2+2x-15)^2}$$

   Recall that critical numbers are the values of the variable where the derivative is zero or does not exist. Since the derivative is a rational function, we know that the derivative will not exist at any value of $x$ that causes the denominator to be zero, and the derivative will equal zero at any value of $x$ in the domain that makes the numerator equal zero.

   So we need to set both the numerator and the denominator equal to zero and solve for $x$. Note that the exponent on the expression in the denominator does not affect where the denominator is zero. So we have the following two equations:

   \begin{align} x^2-2x+13=0 &\Rightarrow x=\frac{2\pm\sqrt{-48}}{2} = 1\pm 2\sqrt{3}i\\ x^2+2x-15=(x+5)(x-3)=0 &\Rightarrow x=-5,3\\ \end{align}

   Recall that critical numbers for us are only real numbers, not complex numbers. So the first equation doesn't contribute any critical numbers.

   Also recall that a number will only be a critical number if it is in the domain of the function (not the derivative, but the original function). In this problem, we have found two values of $x$ where the derivative doesn't exist. But the function also does not exist at these values, so neither of them will be critical numbers.

   Therefore this function has no critical numbers.

9. $r(y)=\sqrt[5]{y^2-6y}$

   Click for solution

   First we need to find the derivative. We'll use the chain rule here: $$r'(y) = \frac15(y^2-6y)^{-\frac45}(2y-6) = \frac{2y-6}{5(y^2-6y)^{\frac45}}$$

   We rewrote the function as a fraction rather than use a negative exponent in order to make the next step easier.

   Recall that critical numbers are the values of the variable where the derivative is zero or does not exist. Since the derivative is a rational function, we know that the derivative will not exist at any value of $y$ that causes the denominator to be zero, and the derivative will equal zero at any value of $y$ in the domain that makes the numerator equal zero.

   So we need to set both the numerator and the denominator equal to zero and solve for $y$. Note that the exponent on the expression in the denominator does not affect where the denominator is zero. So we have the following two equations:

   \begin{align} 2y-6 = 0 &\Rightarrow y=3\\ y^2-6y=y(y-6)=0 &\Rightarrow y=0,6\\ \end{align}

   This process yields three values of $y$; $y=3$ is the value at which the derivative is zero and the other two are the values that cause the derivative to be undefined. Note that all three values are in the domain of the original function, so they are all critical numbers.

   Therefore, we have three critical numbers for this function:

   $$\require{bbox} \bbox[2pt,border:2px solid green]{y=0,3,6}$$

10. $h(t)=15-(3-t)[t^2-8t+7]^{\frac13}$

    Click for solution

    First we need to find the derivative.

    \begin{align} h'(t) &= [t^2-8t+7]^{\frac13}-(3-t)\left(\frac13\right)(2t-8)[t^2-8t+7]^{-\frac23}\\ &= [t^2-8t+7]^{\frac13}-\frac{(3-t)(2t-8)}{3[t^2-8t+7]^{\frac23}}\\ &= \frac{3(t^2-8t+7)-(3-t)(2t-8)}{3(t^2-8t+7)^{\frac23}}\\ &= \frac{5t^2-38t+45}{3(t^2-8t+7)^{\frac23}}\\ \end{align}

    After differentiating, we simplified as much as possible, including moving the expression with the negative exponent to the denominator. This will help considerably with the next step.

    The derivative is now a rational expression. Therefore, the derivative is zero where its numerator is zero and is undefined where its denominator is zero. All we need to do then is set the numerator and the denominator equal to zero and solve for $t$. Note that the multiple and the exponent on the denominator will not affect where the denominator is zero, so we can ignore it. So we just need to solve the following two equations:

    \begin{align} 5t^2-38t+45=0 &\Rightarrow t=\frac{38\pm\sqrt{544}}{10}=\frac{19\pm2\sqrt{34}}{5}\\ t^2-8t+7=0 &\Rightarrow (t-7)(t-1)=0\\ &\Rightarrow t=1,7\\ \end{align}

    This process yields four values of $t$; $t=\frac{19\pm2\sqrt{34}}{5}$ are the values at which the derivative is zero and the other two are the values that cause the derivative to be undefined. Note that all four values are in the domain of the original function, so they are all critical numbers.

    Therefore, we have four critical numbers for this function:

    $$\require{bbox} \bbox[2pt,border:2px solid green]{t=\frac{19\pm2\sqrt{34}}{5}, 1, 7}$$

11. $s(z)=4\cos(z)-z$

    Click for solution

    First let's find the derivative. $$s'(z)=-4\sin(z)-1$$

    Critical numbers are the values of the variable where the derivative is zero or does not exist. This derivative exists everywhere, so we just need to determine where it is zero.

    \begin{align} -4\sin(z)-1=0 &\Rightarrow \sin(z) = -\frac14\\ &\Rightarrow z=\arcsin\left(-\frac14\right)\approx -0.2527\\ \end{align}

    This is the answer that a calculator gives for $\arcsin\left(-\frac14\right)$. We can use this value or the equivalent positive angle: $z\approx 2\pi - 0.2527\approx 6.0305$. We could use either the negative or the positive value; we will use the positive value to find the other values of $z$.

    A look at the unit circle gives us a second solution: $\pi + 0.2527 \approx 3.3943.$

    Using these two solutions, we can now construct all possible solutions to the equation. Hence all the critical numbers of the function are

    $$\require{bbox} \bbox[2pt,border:2px solid green]{\begin{align} z &\approx 6.0305 + 2\pi n\\ z &\approx 3.3943 +2\pi n \hspace{.25in} n=0,\pm1,\pm2,...\\ \end{align} }$$

12. $f(y)=\sin\left(\displaystyle\frac{y}{3}\right) +\displaystyle\frac{2y}{9}$

    Click for solution

    Using the chain rule, we find the derivative is $$f'(y) = \frac13\cos\left(\frac{y}{3}\right) +\frac29$$

    Critical numbers are the values of the variable where the derivative is zero or does not exist. This derivative exists everywhere, so we just need to determine where it is zero.

    \begin{align} \frac13\cos\left(\frac{y}{3}\right)+\frac29 = 0 &\Rightarrow \cos\left(\frac{y}{3}\right) = -\frac23\\ &\Rightarrow \frac{y}{3} = \arccos\left(-\frac23\right) \approx 2.3005\\ \end{align}

    A look at the unit circle gives us a second solution of approximately $-2.3005$ or equivalently, $2\pi -2.3005\approx 3.9827$. These values are equally correct and you can use either. We will use the positive values to construct the rest of the solutions.

    All possible solutions to $\cos\left(\frac{y}{3}\right)=-\frac23$ are then

    \begin{align} \frac{y}{3} &\approx 2.3005 +2\pi n\\ \frac{y}{3} &\approx 3.9827 + 2\pi n \hspace{.25in} n=0,\pm1, \pm2, ...\\ \end{align}

    Finally, solving for $y$ gives us all of the critical numbers of the function.

    $$\require{bbox} \bbox[2pt,border:2px solid green]{ \begin{align} y &\approx 6.9015 +6\pi n\\ y &\approx 11.9481 +6\pi n \hspace{.25in} n=0, \pm1, \pm2, ...\ \end{align} }$$

13. $V(t)=\sin^2(3t)+1$

    Click for solution

    First we need to find the derivative. $$V'(t) = 6\sin(3t)\cos(3t)$$

    Critical numbers are the values of the variable where the derivative is zero or does not exist. This derivative exists everywhere, so we just need to determine where it is zero.

    $$6\sin(3t)\cos(3t)=0 \Rightarrow \sin(3t) = 0 \text{ or} \cos(3t) = 0$$

    So we need to solve these two trigonometric equations.

    Using the unit circle, we see that sine is zero at zero and at $\pi$. All solutions to $\sin(3t)=0$ are then

    \begin{align} 3t = 0 + 2\pi n &\Rightarrow t=\frac23 \pi n\\ 3t = \pi + 2\pi n &\Rightarrow t=\frac13 \pi +\frac23 \pi n\hspace{.25in} n=0,\pm1,\pm2,...\\ \end{align}

    Using the unit circle again, we see that cosine is zero at $\frac{\pi}{2}$ and $\frac{3\pi}{2}$. So all solutions to $\cos(3t)=0$ are

    \begin{align} 3t = \frac{\pi}{2} + 2\pi n &\Rightarrow t=\frac{\pi}{6}+\frac23 \pi n\\ 3t = \frac{3\pi}{2} + 2\pi n &\Rightarrow t=\frac{\pi}{2} +\frac23 \pi n\hspace{.25in} n=0,\pm1,\pm2,...\\ \end{align}

    Therefore, the critical numbers of the function are

    $$\require{bbox} \bbox[2pt,border:2px solid green]{t=\frac23 \pi n, t=\frac13 \pi +\frac23\pi n, t=\frac{\pi}{6}+\frac23\pi n, t=\frac{\pi}{2}+\frac23\pi n \hspace{.25in} n=0,\pm1,\pm2, ...}$$

14. $f(x)=5x\text{e}^{9-2x}$

    Click for solution

    First we will use the product rule to find the derivative. $$f'(x) = 5\text{e}^{9-2x} +5x(-2)\text{e}^{9-2x} = 5\text{e}^{9-2x}(1-2x)$$

    Writing the derivative in factored form helps with the next step. Notice that the derivative exists everywhere, so we only need to determine where the derivative is zero. Notice also that because exponential functions are never zero, the derivative will only be zero if $$1-2x=0 \Rightarrow x=\frac12$$

    So there is a single critical number for this function: $$\require{bbox} \bbox[2pt,border:2px solid green]{x=\frac12}$$

15. $g(w)=\text{e}^{w^3-2w^2-7w}$

    Click for solution

    First we will find the derivative of this function and write it in factored form. $$g'(w) = (3w^2-4w-7)\text{e}^{w^3-2x^2-7w} = (3w-7)(w+1)\text{e}^{w^3-2w^2-7w}$$

    Critical numbers are the values of the variable where the derivative is zero or does not exist. This derivative exists everywhere, so we just need to determine where it is zero. Notice in addition that because exponential functions are never zero, the derivative will only be zero when

    $$(3w-7)(w+1)=0\Rightarrow w=\frac73, -1$$

    So we have two critical numbers for this function:

    $$\require{bbox} \bbox[2pt,border:2px solid green]{w=-1,\frac73}$$

16. $R(x)=\ln(x^2+4x+14)$

    Click for solution

    First we need to find the derivative. $$R'(x)=\frac{2x+4}{x^2+4x+14}$$

    Recall that critical numbers are the values of the variable where the derivative is zero or does not exist. Since the derivative is a rational function, we know that the derivative will equal zero at any value of $x$ in the domain that makes the numerator equal zero.

    We also know that the derivative will not exist at any value of $x$ that makes the denominator equal to zero. However, for this derivative, we notice that its denominator is the polynomial inside of the natural logarithm in the original function. Any $x$ value that makes this polynomial zero will be a value where the function $R(x)$ does not exist. Since these won't be critical numbers, we can skip setting the denominator equal to zero.

    Setting the numerator equal to zero gives

    $$2x+4=0\Rightarrow x=-2$$

    Since the original function is a logarithm, we should check that $-2$ is in the domain of $R(x)$: $$R(-2)=\ln 10$$ which does exist. Therefore, $$\require{bbox} \bbox[2pt,border:2px solid green]{x=-2}$$ is the only critical number for this function.

17. $A(t)=3t-7\ln(8t+2)$

    Click for solution

    First we need to find the derivative.

    $$A'(t) = 3-7\left(\frac{8}{8t+2}\right) = 3-\frac{56}{8t+2} = \frac{24t-50}{8t+2}$$

    Recall that critical numbers are the values of the variable where the derivative is zero or does not exist. Since the derivative is a rational function, we know that the derivative will equal zero at any value of $t$ in the domain that makes the numerator equal zero.

    We also know that the derivative will not exist at any value of $t$ that makes the denominator equal to zero. However, for this derivative, we notice that its denominator is the polynomial inside of the natural logarithm in the original function. Any $t$ value that makes this polynomial zero will be a value where the function $A(t)$ does not exist. Since these won't be critical numbers, we can skip setting the denominator equal to zero.

    Setting the numerator equal to zero gives $$24t-50=0\Rightarrow t=\frac{25}{12}$$

    Since the original function is a logarithm, we should check that $\frac{25}{12}$ is in the domain of $A(t)$: $$A\left(\frac{25}{12}\right)=\frac{75}{12}-7\ln\left(\frac{65}{3}\right)$$ which is defined, since $\frac{65}{3}>0$. Therefore, $$\require{bbox} \bbox[2pt,border:2px solid green]{t=\frac{25}{12}}$$ is the only critical number for this function.

Assignment Problems

For problems 1 - 40, determine the critical numbers of the function.

$R(x)=8x^3-18x^2-240x+2$

$f(z)=2z^4-16z^3+20z^2-7$

$g(z)=8-12z^5-25z^6+\frac{90}{7}z^7$

$P(w)=w^3-4w^2-7w-1$

$A(t)=7t^3-3t^2+t-15$

$a(t)=4-2t^2-6t^3-3t^4$

$h(v) = v^5+v^4+10v^3-15$

$g(z)=(z-3)^5(2z+1)^4$

$R(q)=(q+2)^4(q^2-8)^2$

$f(t)=(t-2)^3(t^2+1)^2$

$f(w)=\displaystyle\frac{w^2+2w+1}{3w-5}$

$h(t)=\displaystyle\frac{3-4t}{t^2+1}$

$R(y)=\displaystyle\frac{y^2-y}{y^2+3y+8}$

$Y(x)=\sqrt[3]{x-7}$

$f(t)=(t^3-25t)^{\frac23}$

$h(x)=\sqrt[5]{x}(2x+8)^2$

$Q(w)=(6-w^2)\sqrt[3]{w^2-4}$

$Q(t)=7\sin\left(\displaystyle\frac{t}{4}\right)-2$

$g(x)=3\cos(2x)-5x$

$f(x)=7\cos(x)+2x$

$h(t)=6\sin(2t)+12t$

$w(z)=\cos^3\left(\displaystyle\frac{z}{5}\right)$

$U(z)=\tan(z)-4z$

$h(x)=x\cos(x)-\sin(x)$

$h(x)=2\cos(x)-\cos(2x)$

$f(w)=\cos^2(w)-\cos^4(w)$

$F(w)=\text{e}^{14w+3}$

$g(z)=z^2\text{e}^{1-z}$

$A(x)=(3-2x)\text{e}^{x^2}$

$P(t)=(6t+1)\text{e}^{8t-t^2}$

$f(x)=\text{e}^{3+x^2}-\text{e}^{2x^2-4}$

$f(z)=\text{e}^{z^2-4z}+\text{e}^{8z-2z^2}$

$h(y)=\text{e}^{6y^3-8y^2}$

$g(t)=\text{e}^{2t^3+4t^2-t}$

$Z(t)=\ln(t^2+t+3)$

$G(r)=r-\ln(r^2+1)$

$A(z)=2-6z+\ln(8z+1)$

$f(x)=x-4\ln(x^2+x+2)$

$g(x)=\ln(4x+2)-\ln(x+4)$

$h(t)=\ln(t^2-t+1)+\ln(4-t)$



1. The graph of a function $f(x)$ is shown below. Use the graph to estimate the critical numbers of the function.
   

2. The graph of a function $f(x)$ is shown below. Use the graph to estimate the critical numbers of the function.
   

3. The graph of a function $f(x)$ is shown below. Use the graph to estimate the critical numbers of the function.