Exercises
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Exercises
On what intervals are the following functions increasing vs. decreasing?
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$f(x) = 4x^3 + 2x - 1$Find the absolute extrema of the following functions on the given intervals.
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Find where the local extrema of the following functions occur and what kind they are.
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Explain why none of the following functions have and absolute maximum or absolute minimum on their natural domains.
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Determine which of the following functions satisfy the requirements of the Mean Value Theorem on the given intervals. If they don't then explain why and, if they do, find the value the theorem says has to exist.
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Explorations
- Show that if $f(x)$ is increasing on the interval $[a, b]$ then $f(x) + h$ is also increasing on that interval for any real number $h$.
- Show that if $f(x)$ is increasing on the interval $[a, b]$ then $f(x + h)$ is also increasing on the interval $[a-h, b-h]$.
- Rolle's Theorem is a special case of the Mean Value Theorem that says if a function is continuous on a closed interval, $[a, b]$, and differentiable on the open interval, $(a, b)$, and $f(a) = f(b)$ then there exists a value, $c$ in $[a, b]$ such that $f'(c) = 0$. Explain why this is a special case of the Mean Value Theorem.
- Use Rolle's Theorem to show that a third degree polynomial can have at most three roots.
- The arithmetic mean (or average) of two numbers, $a$ and $b$ is $\frac{a + b}{2}$. Show that the value guaranteed by the mean value theorem for the function $f(x) = x^2$ on the interval $[a, b]$ is the arithmetic mean of the end points.
- Assume that $f$ and $g$ are continuous on $[a, b]$ and differentiable on $(a, b)$. If $f(a) = g(a)$ and $f(b) = g(b)$ then show that there exists a point on $[a, b]$ where the tangent lines to the two functions at that point are parallel.
- Show that $|\sin a - \sin b| \le |a - b|$ for all $a$ and $b$.
- Let $f$ be a function with $f(x) > 0$ for all $x$ and let $g = 1/f$. If $f$ is increasing on an interval, is $g$ also increasing on the interval? If $f$ has a local maximum (or minimum) at a point, does $g$ also have one at that point?
Icons courtesy of icons8.com
Where is the function $f(x) = x^2$ increasing vs. decreasing?
First, we need to find the critical values which are the points where the first derivative is equal to 0.
$$f'(x) = 2x$$ $$2x = 0$$ $$x = 0$$So we only have one critical value, i.e. one value where the sign of the derivative can potentiall change sign.
| Interval | Test Point (c) | $f'(c)$ |
| $(-\infty, 0)$ | -1 | $f'(-1) = -2 < 0$ |
| $(0, \infty)$ | 1 | $f'(1) = 2 > 0$ |
Therefore, the function is decreasing on the interval $(-\infty, 0)$ and increasing on $(0, \infty)$.
Where is the function $f(x) = 4x^3 + 2x - 1$ increasing vs. decreasing?
Remember that it's possible for a function to be strictly increasing or strictly decreasing, i.e. it can be increasing (or decreasing) everywhere.
Where is the function $f(x) = 4x^3 + 2x - 1$ increasing vs. decreasing?
First, we need to find the critical values which are the points where the first derivative is equal to 0.
$$f'(x) = 4x^2 + 2$$ $$4x^2 + 2 = 0$$ $$4x^2 = -2$$There are no values where that's true so we only have one interval to consider.
| Interval | Test Point (c) | $f'(c)$ |
| $(-\infty, \infty)$ | 0 | $f'(0) = 2 < 0$ |
This means that the derivative is positive everywhere so the function must be increasing everywhere.
Where is the function $f(x) = x\ln x$ increasing vs. decreasing?
First, we need to find the critical values which are the points where the first derivative is equal to 0.
$$f'(x) = \ln x \cdot \frac{d}{dx} x + x\frac{d}{dx}\ln x$$ $$f'(x) = \ln x + x\frac{1}{x}$$ $$f'(x) = \ln x + 1$$ $$\ln x + 1 = 0$$ $$\ln x = -1$$ $$x=e^{-1}$$When setting up the intevals, you have to keep the domain of the original function in mind. Since you can't take the logarithm of a negative number, the domain is $(0, \infty)$ so the function can only be increasing or decreasing on a subset of that interval. Our critical value divides it into two parts.
| Interval | Test Point (c) | $f'(c)$ |
| $(0, e^{-1})$ | e^{-2} | $f'(e^{-2}) = \ln e^{-2} + 1 = -2 + 1 = -1< 0$ |
| $(e^{-1}, \infty)$ | e^{2} | $f'(e^{2}) = \ln e^{2} + 1 = 2 + 1 = 3> 0$ |
This means that the function is increasing on $(e^{-1}, \infty)$ an decreasing on $(0, e^{-1})$.
Find the absolute extrema of the $f(x) = \sqrt{x-2}$ on the interval $1 \le x \le 10$.
The method here is to find all the critical values, i.e. points where the derivative equals 0, that are on the given interval. Then we'll evaluate the original function at the critical values and the interval's end points and take the largest and smallest as the answers.
First notice that the domain of the function is $[2, \infty)$. We don't want to consider points that aren't on that interval so I'm going to change the given interval from $1 \le x \le 10$ to $2 \le x \le 10$. This is important since we'll need to be sure to test the actual/practical end point of the interval.
$$f(x) = (x-2)^{1/2}$$ $$f'(x) = \frac{1}{2}(x-2)^{-1/2} = \frac{1}{2(x-2)^{1/2}}$$The numerator is never equal to 0 and the denominator is equal to 0 at $x = 2$ so that's our only critical value and it is on the given interval so we have three points to test.
| Value (c) | $f(c)$ |
| 2 | $f(2) = \sqrt{2 - 2} = 0$ |
| 10 | $f(10) = \sqrt{10-2} = \sqrt{8}$ |
This means that the function has a minimum value of 0 at $x=2$ and a maximum of $\sqrt{8}$ at $x=10$.
Notice how, even if we had tested $x=-1$ we still would haven't considered it as a result since the derivative is undefined there. This is good but you can't count on the "real" end point also being a critical value so you should always start by adjusting the interval the way that I did in the first paragraph.
Find the absolute extrema of the $f(x) = \csc x$ on the interval $\pi/3 \le x \le 3\pi/4$.
The cosecant is defined for all the values on $\pi/4 \le x \le 3\pi/4$ so we don't have to worry about changing the given interval.
$$f(x) = \csc x$$ $$f'(x) = -\csc x\cot x$$$\csc x$ is never equal to 0 and $\cot x = 0$ only when $x=\pi/2$. (I'm only interested in values that are on the given interval.)
| Value (c) | $f(c)$ |
| $\pi/3$ | $f(\pi/3) = \csc(\pi/3) = 2\sqrt{3}/3 \approx 1.155$ |
| $\pi/2$ | $f(\pi/2) = \csc(\pi/2) = 1$ |
| $3\pi/4$ | $f(3\pi/4) = \csc(3\pi/4) = \sqrt{2} \approx 1.414 $ |
This means that the function has an absolute minimum value of 1 at $x=\pi/2$ and an absolute maximum of $\sqrt{2}$ at $x=3\pi/4$.
Find the absolute extrema of the $f(x) = -2 x^{1/3}$ on the interval $-1 \le x \le 2$.
The cube root function is defined everywhere so we don't have to worry about changing things because of the domain.
$$f(x) = -2 x^{1/3}$$ $$f'(x) = -\frac{2}{3} x^{-2/3}$$The denominator part, $x^{-2/3}$, is 0 when $x=0$ so that will be our only critical value.
| Value (c) | $f(c)$ |
| -1 | $f(-1) = 2$ |
| 0 | $f(0) = 0$ |
| 2 | $f(2) = -2 \cdot 2^{1/3} \approx -2.520 $ |
This means that the function has an absolute minimum value of -2 \cdot 2^{1/3} at $x=2$ and an absolute maximum of 2 at $x=-1$.
Find where the local extrema of the function $f(x) = x^3 + 9x^2 - 4x + 1$ occur and what kind they are.
We need to start by finding the critical values where the function is defined then look to see if it changes from increasing to decreasing or decreasing to increasing on each side.
$$f'(x) = 3x^2 + 18x - 4$$ $$3x^2 + 18x - 4 = 0$$This doesn't look like it's going to factor so we'll have to find the solutions using the Quadratic Formula.
$$x=\frac{-18\pm\sqrt{18^2-4(3)(-4)}}{2\cdot 3}$$ $$x=\frac{-18\pm\sqrt{372}}{6}$$ $$x=\frac{-18\pm2\sqrt{93}}{6}$$ $$x=\frac{-9\pm\sqrt{93}}{3}$$ $$x\approx-6.215, 0.215$$| Interval | $\left(\infty, \frac{-9-\sqrt{93}}{3}\right)$ | $\left(\frac{-9-\sqrt{93}}{3}, \frac{-9+\sqrt{93}}{3}\right)$ | $\left(\frac{-9-\sqrt{93}}{3}, \infty\right)$ |
| Test Value (c) | -7 | 0 | 1 |
| $f'(c)$ | $f'(-7)=17$ | $f'(0)=-4$ | $f'(1)=17$ |
The 17 and -4 tell us that the function increases up to $\frac{-9-\sqrt{93}}{3}$ then decreases away from it so that point is a relative maximum then it decreases to $\frac{-9+\sqrt{93}}{3}$ and increases away from it so that point is a relative minimum.
Find where the local extrema of the function $f(x) = x^3 + 9x^2 - 4x + 1$ occur and what kind they are.
We need to start by finding the critical values where the function is defined then look to see if it changes from increasing to decreasing or decreasing to increasing on each side.
$$\begin{aligned} f'(x) &= \ln x\frac{d}{dx}x^2 + x^2 \frac{d}{dx}\ln x \\ &= 2x\ln x + x^2 \frac{1}{x} \\ &= 2x\ln x + x \end{aligned}$$ $$2x\ln x + 2 = 0$$ $$x(2\ln x + 1) = 0$$| $$x=0$$ | $$2\ln x + 1 = 0$$ $$\ln x = -1/2$$ $$x = e^{-1/2}$$ |
Since the function isn't defined for negative values, our first interval will start at 0.
| Interval | $\left(0, e^{-1/2}\right)$ | $\left(e^{-1/2}, \infty\right)$ |
| Test Value (c) | $e^{-1}$ | 1 |
| $f'(c)$ | $\begin{aligned}f'(e^{-1})&=2e^{-1}\ln e^{-1} + e^{-1} \\ &= -2e^{-1} + e^{-1} \\ &= -e^{-1}\end{aligned}$ | $f'(1)=17$ |
The derivative goes from negative to positive as the function passes through $e^{-1/2}$ so it decreases to that point then increases away from it making the point a relative minimum.
Find where the local extrema of the function $f(x) = x^{-5/3}$ occur and what kind they are.
We need to start by finding the critical values where the function is defined then look to see if it changes from increasing to decreasing or decreasing to increasing on each side.
$$f'(x) = -\frac{5}{3}x^{-7/3} = -\frac{5}{3x^{7/3}}$$The numerator is never equal to 0 but the denominator equals 0 when $x=0$ so that's our only critical value but, since the function is undefined there, it can't be a relative maximum or minimum. Since that was the only critical value, our conclusion is that the function doesn't have any extrema..
Find where the local extrema of the function $f(x) = \begin{cases}3x - 1,&x < 2\\-2x+1, & x \ge 2\end{cases}$ occur and what kind they are.
For a piecewise defined function, we need to look at each of the parts separately. For $x < 2$, $f'(x) = 3$ so the only possible extrema is at the end point but there isn't one on this side since the interval doesn't actually include 2.
For $x \ge 2$, $f'(x) = -2$ so the function is decreasing making the left end point, $(2, -3), a potential local maximum.$,
However, if we look at the other side, $\lim\limits_{x\to 2^-}f(x) = 3 \cdot 2 - 1 = 6$ which is greater than -3. This means that, every neighborhood of 2 contains some points that are greater than -3 (all the points on the left side) so it can't be a maximum so the function has no extrema.
Explain why the function $f(x) = \cos x$ has no absolute maxima or absolute minima on their natural domains.
$$f'(x) = -\sin x$$
This is equal to 0 at $x=k\pi$ for any integer $k$. For any even value of $k$, i.e. when $x=...,-4\pi, -2\pi, 0, 2\pi, 4\pi, ...$ the derivative goes from positive to negative as it passes through each point so we could expect those points to be relative maxima. However, if $k$ is odd then
$$f(k\pi) = \cos(k\pi) = 1$$This is a problem because all the function values are the same. If you look closely at the definition of an absolute extrema, it says the function value has to be strictly greater than every other function value. This doesn't allow for the case where two values could be equal. In other words, an abolute maximum (or minimum) has to be unique. Since our value occurs at multiple places, this function doesn't have an absolute maximum.
The function also doesn't have an absolute minimum for the same reason. THose values would occur when $k$ is odd but $f(k\pi) = -1$ at all of those points.
Determine if the function $f(x) = x^{4/3}$ satisfies the requirements of the Mean Value Theorem on the interval $[1, 8]$. If they don't then explain why and, if they do, find the value the theorem says has to exist.
This function is continuous everywhere and its derivative, $f'(x) = \frac{4}{3}x^{1/3}$$ is also defined everywhere so the Theorem's requirements are met.
The theorem says that there exists a $c$ on the interval such that
$$\frac{f(1)-f(8)}{1-8} = f'(c)$$ $$\frac{1 - 16}{-7} = \frac{4}{3}c^{1/3}$$ $$\frac{15}{7} = \frac{4}{3}c^{1/3}$$ $$\frac{45}{28} = c^{1/3}$$ $$c = \left(\frac{45}{28}\right)^3=\frac{91125}{21952}$$That's approximately equal to 4.15 so it is on the given interval.
Determine if the function $f(x) = an x$ satisfies the requirements of the Mean Value Theorem on the interval $[0, \pi]$. If they don't then explain why and, if they do, find the value the theorem says has to exist.
The tangent function is undefined at $x=\pi/2$ which is on the interval $[0, \pi]$. Since it's undefined at a point on the interval it can't be continuous (or differentiable) at that point so the Mean Value Theorem doesn't apply.
Determine if the function $f(x) = \begin{cases}3x+1,&x<2\\2x^2-x+1,&x\ge 2\end{cases}$ satisfies the requirements of the Mean Value Theorem on the interval $[-2, 2]$. If they don't then explain why and, if they do, find the value the theorem says has to exist.
$$\lim\limits_{x\to2^-} = 3\cdot 2 + 1 = 7$$ $$\lim\limits_{x\to2^+} = 2\cdot 2^2 - 2 + 1 = 7$$ $$f(2) = 7$$
Since the left and right limits are equal, we have
$$\lim\limits_{x\to2}f(x) = f(2)$$So the function is continuous at $x=2$. However, to the left of 2, we have $$f'(x) = 3$$, so
$$\lim\limits_{x\to2^-}f'(x) = 3$$However, on the right side, we have $f'(x) = 4x-1$ so
$$\lim\limits_{x\to2^+}f'(x) = 4\cdot 2 - 1 = 7 \neq 3$$Since the left and right hadn limits don't match, the function isn't differentiable at $x=2$ so the Mean Value Theorem's conditions aren't met.
If I wanted to write the derivative of the function, it would be
$$f(x) = \begin{cases}3,&x<2\\4x-1,&x > 2\end{cases}$$Notice that 2 isn't included in either of the two intervals.
Show that if $f(x)$ is increasing on the interval $[a, b]$ then $f(x) + h$ is also increasing on that interval for any real number $h$.
Let $g(x) = f(x) + h$. If $f(x)$ is increasing on $[a, b]$, that means that $f'(c) > 0$ for every $c$ on the interval. Now think about $g'(x)$.
$$g'(x) = \frac{d}{dx}(f(x) + h) = f'(x)$$More specifically, $g'(c) = f'(c)$ for every $c$ on $[a, b]$ so if $f'(c) > 0$ for every point on the interval then $g'(c)$ is also so $g$ is also always increasing on $[a, b]$.
Rolle's Theorem is a special case of the Mean Value Theorem that says if a function is continuous on a closed interval, $[a, b]$, and differentiable on the open interval, $(a, b)$, and $f(a) = f(b)$ then there exists a value, $c$ in $[a, b]$ such that $f'(c) = 0$. Explain why this is a special case of the Mean Value Theorem.
Click on the colored blocks to uncover the text.
First notice that both theorem's have the same assumptions about so those points are both covered.
The Mean Value Theorem says that there exists such that
Rolle's Theorem adds the additional requirement that which means that . If we make that substitution in the Mean Value Theorem's formula, it requires that there exists a $c$ such that
Which is the conclusion of Rolle's Theorem.
Rolle's Theorem is a special case of the Mean Value Theorem that says if a function is continuous on a closed interval, $[a, b]$, and differentiable on the open interval, $(a, b)$, and $f(a) = f(b)$ then there exists a value, $c$ in $[a, b]$ such that $f'(c) = 0$. Explain why this is a special case of the Mean Value Theorem.
First notice that both theorem's have the same assumptions about continuity and differentiability so those points are both covered.
The Mean Value Theorem says that there exists a point $c$ on $[a, b]$ such that
$$\frac{f(b) - f(a)}{b - a} = f'(c)$$Rolle's Theorem adds the additional requirement that $f(a) = f(b)$ which means that $f(a) - f(b) = 0$. If we make that substitution in the Mean Value Theorem's formula, it requires that there exists a $c$ such that
$$f'(c) = \frac{0}{b - a} = 0$$Which is the conclusion of Rolle's Theorem.
Use Rolle's Theorem to show that a third degree polynomial can have at most three roots.
Click on the colored blocks to uncover the text.
Suppose that there's a third degree polynommial with four roots. That means there are distinct values, , such that . Rolle's theorem says that there must be three distinct values that make , one between , one between but this can't happen. The derivative of a third degree polynomial is a polynomial and a second degree polynomial has at most .
Use Rolle's Theorem to show that a third degree polynomial can have at most three roots.
Suppose that there's a third degree polynommial with four roots. That means there are three distinct values, $a < b < c < d$, such that $f(a) = f(b) = f(c) = f(d) = 0$. Rolle's theorem says that there must be three distinct values that make $f'(x) = 0$, one between $a$ and $b$, one between $b$ and $c$ and one between $c$ and $d$ but this can't happen. The derivative of a third degree polynomial is a second degree polynomial and a second degree polynomial has at most two values where it equals 0.
The arithmetic mean (or average) of two numbers, $a$ and $b$ is $\frac{a + b}{2}$. Show that the value guaranteed by the mean value theorem for the function $f(x) = x^2$ on the interval $[a, b]$ is the arithmetic mean of the end points.
$f(x) = x^2$ is continuous and diferentiable everywhere so the requirements of the Mean Value Theorem are met. So now consider an interval $[a, b]$. The Mean Value Theorem says that, on this interval, there exists a point $c$ such that
$$f'(c) = \frac{f(a)-f(b)}{a-b}$$ $$2c = \frac{a^2 - b^2}{a - b}$$ $$2c = \frac{(a-b)(a+b)}{a-b}$$ $$2c = a+b$$ $$c = \frac{a+b}{2}$$Which is the arithmetic mean of the two endpoints.
Show that $|\sin a - \sin b| \le |a - b|$ for all $a$ and $b$.
We're going to use the Mean Value Theorem here with the function $f(x) = \sin x$ so, first, notice that that function so the Mean Value Theorem applies on any interval $[a, b]$. According to that theorem, there exists a point, $c$, on that interval such that
or
$$\sin a - \sin b = (a-b)\cos c$$At this point, I don't know what $\cos c$ is but I know that it's less than or equal to since that's the maximum value of the cosine function. This means that
Show that $|\sin a - \sin b| \le |a - b|$ for all $a$ and $b$.
We're going to use the Mean Value Theorem here with the function $f(x) = \sin x$ so, first, notice that that function is continuous and differentiable everywhere so the Mean Value Theorem applies on any interval $[a, b]$. According to that theorem, there exists a point, $c$, on that interval such that
$$f'(c) = \frac{f(a) - f(b)}{a - b}$$or
$$\sin a - \sin b = (a-b)\cos c$$ $$|\sin a - \sin b| = |(a-b)\cos c|$$ $$|\sin a - \sin b| = |a-b||\cos c|$$At this point, I don't know what $\cos c$ is but I know that it's less than or equal to 1 since that's the maximum value of the cosine function. This means that
$$|\sin a - \sin b| = |a-b||\cos c| \le |a-b|\cdot 1 = |a-b|$$ $$|\sin a - \sin b| \le |a-b|$$Let $f$ be a function with $f(x) > 0$ for all $x$ and let $g = 1/f$. If $f$ is increasing on an interval, is $g$ also increasing on the interval? If $f$ has a local maximum (or minimum) at a point, does $g$ also have one at that point?
First notice that
$$g(x) = \{f(x)\}^{-1}$$so
$$g'(x) = -\{f(x)\}^{-2} \cdot f'(x) = -\frac{f'(x)}{f^2(x)}$$Now, if $f$ is increasing on an interval, that means that both $f(x) > 0$ and $f'(x) > 0$ which means that $$g'(x) = -\frac{f'(x)}{f^2(x)} = -\frac{\text{a positive number}}{\text{a positive number}} < 0$$
Since $g'(x)$ is negative everywhere on the interval, it must actually be decreasing on it.
Now suppose $f(x)$ has a local maximum at a point $c$, i.e. $f'(c) = 0$. Using the formula we calculated earlier for $g'(x)$, this means that
$$g'(c) = -\frac{f'(c)}{f^2(c)} = -\frac{0}{f^2(c)} = 0$$So $c$ is a critical value for $g$ but we still need to see if there really is an extremum there. Since we're assuming that $c$ is a local maximum, that means that $f'(x)>0$ on the left side of $c$ and $f'(x) < 0$ on the right side of $c$. So what does that tell use about $g'$ around that point? On the left side, we would have
$$g'(x) = -\frac{f'(x)}{f^2(x)} = -\frac{\text{a positive number}}{\text{a positive number}} < 0$$and, on the right side,
$$g'(x) = -\frac{f'(x)}{f^2(x)} = -\frac{\text{a negative number}}{\text{a positive number}} > 0$$That means that the graph decreases to $c$ and then increases away from it so this point must be a local minimum.
