# Math 141, Lesson 28: The Derivative as a Function II

The lesson page, with interactive tools and R in the browser: https://tangentspace.org/lessons/28/

This lesson is marked "Updates coming": parts of it are still being written.

Reading: MMAC Ch 4.2, pp 369-378

Practice: Optional HW 28 **[Q?? and Q??]**.

Interactive tools: [Linked Derivative Graphs](https://tangentspace.org/tools/linked-derivatives/)

## Objectives

1. Describe *differentiable* and identify at what inputs a function is or is not differentiable.
2. Given a function, use R to approximate and plot its first and higher-order derivatives.
3. Given a first derivative, identify where the function is increasing or decreasing and identify local extrema.
4. Given a first or second derivative, determine concavity and identify inflection points.

## 1. Differentiability

*Objective.* Describe *differentiable* and identify at what inputs a function is or is not differentiable.

**Definition 1** (Differentiable).

A function $f(x)$ is *differentiable* at $x = a$ if and only if the derivative $f'(a)$ exists. A function $f(x)$ is differentiable on an interval if and only if $f(x)$ is differentiable at every input in the interval.

The function $f(x) = |x|$ is not differentiable at $x = 0$, because the derivative $f'(0)$ does not exist. Its average rates of change approach 1 from the right and $-1$ from the left ([Lesson 26](https://tangentspace.org/lessons/26/) Notes, Example 11).

Graphical features that correspond to a function being nondifferentiable are sharp points, including corners and cusps, as well as breaks in the graph, including holes, jumps, and vertical asymptotes. Each graph below is the graph $y = f(x)$ of a function $f$ with one feature at $x = 0$.

**Corner**

![Figure](https://tangentspace.org/lessons/fig/1e8fadf56536.svg)

The slopes on the two sides of $x = 0$ are different numbers, $-1$ and 1.

**Cusp**

![Figure](https://tangentspace.org/lessons/fig/af20e635ce9e.svg)

The graph comes to a sharp point, and near $x = 0$ the slopes grow without bound in size.

**Hole**

![Figure](https://tangentspace.org/lessons/fig/ad5353aef6ac.svg)

The derivative $f'(0)$ does not exist, because $f(0)$ is undefined and the limit definition of the derivative uses $f(0)$.

**Jump**

![Figure](https://tangentspace.org/lessons/fig/15ab57a6e724.svg)

The filled dot is $f(0)$, and the outputs from the left approach a lower value. Over a short interval $[x, 0]$ on the left, the change in output stays near the height of the jump while the interval’s length shrinks toward 0. Their quotient, the average rate of change, grows without bound.

**Vertical asymptote**

![Figure](https://tangentspace.org/lessons/fig/011c122a2c40.svg)

The function has no output at $x = 0$, and near $x = 0$ its outputs grow without bound in size.

### Example 1

Sketch the graph of the derivative of the function $f(x)$ graphed below, and identify the inputs $x$ with $-6 < x < 9$ where $f$ is not differentiable.

![Figure](https://tangentspace.org/lessons/fig/b0da0adf372b.svg)

**Solution.**

By the definition of differentiable, $f$ is differentiable at $x = a$ exactly when $f'(a)$ exists. On each of the five line segments, $f'(x)$ is the slope of the segment:

$$f'(x) = \begin{cases}
1 & -6 < x < -3 \\
0 & -3 < x < 0 \\
-1 & 0 < x < 3 \\
0 & 3 < x < 6 \\
4 & 6 < x < 9
\end{cases}.$$

Each slope is the rise over the run, such as $\frac{6 - (-6)}{9 - 6} = 4$ on the last segment. At $x = -3$, every interval $[x, -3]$ with $-6 \le x < -3$ gives an average rate of change of $1$, and every interval $[-3, x]$ with $-3 < x \le 0$ gives $0$. Because the left and right average rates of change approach different numbers, $f'(-3)$ does not exist. The slopes on the two sides of $x = 0$, $x = 3$, and $x = 6$ also differ. Therefore, $f$ is not differentiable at $x = -3$, $0$, $3$, and $6$. The graph beside this text is the sketch of $f'$: one horizontal segment at the height of each slope, with open dots at the four inputs where $f'$ does not exist.

![Figure](https://tangentspace.org/lessons/fig/155b3c3d57d3.svg)

### Example 2

Identify the inputs $x$ with $-6 < x < 6$ where the function $h(x)$ graphed below is not differentiable, and name the graphical feature at each one.

![Figure](https://tangentspace.org/lessons/fig/1b91452656eb.svg)

**Solution.**

The function $h$ is not differentiable at $x = -3$, $-1$, $1$, and $4$.

- The graph of $h$ has a corner at $x = -3$, because the slope of $h$ is $1$ on the left and $-1$ on the right.
- The graph of $h$ has a hole at $x = -1$, because the outputs approach $1$ from both sides and $h(-1)$ is undefined.
- The graph of $h$ has a jump at $x = 1$, because the outputs approach $-1$ from the left and $2$ from the right.
- The graph of $h$ has a vertical asymptote at $x = 4$, because the outputs grow without bound as $x$ approaches $4$ from the left and fall without bound from the right.

## 2. Higher-Order Derivatives in R

*Objective.* Given a function, use R to approximate and plot its first and higher-order derivatives.

**Definition 2** (Second and higher-order derivatives).

The *second derivative* $f''(x)$ of a function $f(x)$ is the derivative of $f'(x)$. In general, $f^{(1)}(x)$ means $f'(x)$, and for each integer $n$ with $n \ge 2$, the $n$th derivative $f^{(n)}(x)$ of $f(x)$ is the derivative of $f^{(n-1)}(x)$. The derivatives $f^{(n)}(x)$ with $n \ge 2$ are the *higher-order derivatives* of $f(x)$.

The function $f(x) = |x|$ has no second derivative at $x = 0$, because $f'(0)$ does not exist. The derivative $f'(x)$ is $-1$ for $x < 0$ and 1 for $x > 0$, and the derivative of $f'$ at $x = 0$ uses the value $f'(0)$.

Compared with the dimension of $f'(x)$, the dimension of $f''(x)$ has one more factor of $[\text{input}]^{-1}$:

$$[f'(x)] = [\text{output}]\cdot[\text{input}]^{-1},
\qquad
[f''(x)] = [\text{output}]\cdot[\text{input}]^{-2}.$$

The [Lesson 25](https://tangentspace.org/lessons/25/) Notes, section 1, give the units of a rate of a rate, such as an acceleration in m/s$^2$ with dimension $LT^{-2}$.

### Example 3

Use R to approximate and plot the first and second derivatives of

$$f(x) = x^3 - 5x$$

on $-3 \le x \le 3$, and estimate $f'(2)$ and $f''(2)$.

**Solution.**

By the definition of the second derivative, $f''$ is the derivative of $f'$. Define $f$ and the step-0.001 approximation `f_deriv` of $f'$. The function `f_deriv2` applies the same approximation to `f_deriv` and so approximates $f''$. Plot all three on $-3 \le x \le 3$. Used R.

```r
f = makeFun(x^3 - 5*x ~ x)
plotFun(f(x) ~ x, xlim = range(-3, 3), col = "black")
f_deriv = makeFun((f(x + 0.001) - f(x))/0.001 ~ x)
f_deriv2 = makeFun((f_deriv(x + 0.001) - f_deriv(x))/0.001 ~ x)
plotFun(f_deriv(x) ~ x, col = "darkgreen", add = TRUE)
plotFun(f_deriv2(x) ~ x, col = "purple", add = TRUE)
```

In the plot, the black cubic is $f$, the green parabola is $f'$, and the purple line is $f''$. The parabola crosses the $x$-axis at the inputs where the cubic has slope 0, and the line crosses the $x$-axis at $x = 0$, where the parabola has slope 0.

![Figure](https://tangentspace.org/lessons/fig/5dc41df1f6b8.svg)

Evaluate both approximations at $x = 2$. Used R.

```r
f_deriv(2)
#> [1] 7.006001
f_deriv2(2)
#> [1] 12.006
```

The step-0.001 values 7.006 and 12.006 are first estimates. To find $f'(2)$ and $f''(2)$ to three decimals, apply the stopping test of the [Lesson 25](https://tangentspace.org/lessons/25/) Notes, section 2: shrink the step on both sides of $x = 2$ until the third decimal stops changing.

The second expression below repeats the computation of `f_deriv2` with step $h$. Its part `(f(2 + 2*h) - f(2 + h))/h` is the first-derivative estimate at $2 + h$, and `(f(2 + h) - f(2))/h` is the first-derivative estimate at 2. Their difference divided by $h$ is the average rate of change of the first-derivative estimates from 2 to $2 + h$, which approximates $f''(2)$. The three negative entries of `h` are steps to the left of 2. Used R.

```r
h = c(0.0001, 0.00001, 0.000001, -0.0001, -0.00001, -0.000001)
(f(2 + h) - f(2))/h
#> [1] 7.000600 7.000060 7.000006 6.999400 6.999940 6.999994
((f(2 + 2*h) - f(2 + h))/h - (f(2 + h) - f(2))/h)/h
#> [1] 12.00060 12.00004 11.99751 11.99940 11.99997 11.99840
```

The first-derivative estimates round to $7.000$ from both sides at steps 0.00001 and 0.000001. The stopping test is met:

$$f'(2) \approx 7.000.$$

The second-derivative estimates round to $12.000$ from both sides at step 0.00001, and at step 0.000001 they drift to $11.998$. At so small a step, the rounding error in R’s stored outputs of $f$ is divided by $h$ twice in the formula, and the result is large enough to change the third decimal. The stopping test therefore does not confirm $f''(2)$. The step-0.00001 estimates come before that drift, and they give the estimate:

$$f''(2) \approx 12.000.$$

### Example 4

The height $H$ in meters of a ball thrown straight up, $t$ seconds after release, is

$$H(t) = 19.6t - 4.9t^{2}.$$

Use R to approximate $H'(1)$ and $H''(1)$ with step 0.001, give the units of each, and plot $H'$ and $H''$ for $0 \le t \le 4$.

**Solution.**

Define $H$ and the step-0.001 approximation `H_deriv` of $H'$, then `H_deriv2`, which applies the same approximation to `H_deriv`. Evaluate both at $t = 1$, then plot both. Used R.

```r
H = makeFun(19.6*t - 4.9*t^2 ~ t)
H_deriv = makeFun((H(t + 0.001) - H(t))/0.001 ~ t)
H_deriv2 = makeFun((H_deriv(t + 0.001) - H_deriv(t))/0.001 ~ t)
H_deriv(1)
#> [1] 9.7951
H_deriv2(1)
#> [1] -9.8
plotFun(H_deriv(t) ~ t, xlim = range(0, 4), col = "darkgreen")
plotFun(H_deriv2(t) ~ t, col = "purple", add = TRUE)
```

At step 0.001 the error of the estimate of $H'(1)$ shows in the second decimal. Report it to one decimal:

$$H'(1) \approx 9.8, \qquad H''(1) \approx -9.800.$$

The derivative $H'$ is the ball’s velocity, in meters per second. The second derivative $H''$ is the rate of change of the velocity, in meters per second per second, written m/s$^2$, with dimension $LT^{-2}$. One second after release, the height is increasing at about 9.8 m/s, and the velocity is decreasing at about 9.800 m/s$^2$.

In the plot, the green line $H'$ falls from about 19.6 to about $-19.6$. The purple line $H''$ stays at height $-9.8$ at every time.

![Figure](https://tangentspace.org/lessons/fig/ae07cf52cb63.svg)

*Third and higher derivatives in R.* To approximate each further derivative in R, apply the same step-0.001 formula to the approximation of the derivative before it. The line below defines an approximation of the third derivative $f'''(x)$, also written $f^{(3)}(x)$.

```r
f_deriv3 = makeFun((f_deriv2(x + 0.001) - f_deriv2(x))/0.001 ~ x)
```

*Order of the R lines.* The function `f_deriv2` calls `f_deriv`, and `f_deriv` calls `f`. Evaluating or plotting `f_deriv2` before `f_deriv` is defined gives an error:

```r
> f_deriv2(1)
Error in f_deriv(x + 0.001) : could not find function "f_deriv"
```

## 3. Increasing, Decreasing, and Local Extrema

*Objective.* Given a first derivative, identify where the function is increasing or decreasing and identify local extrema.

**Theorem 3** (Derivative and monotonicity).

If $f'(x) > 0$ for every $x$ in an interval, then $f$ is increasing on that interval. If $f'(x) < 0$ for every $x$ in an interval, then $f$ is decreasing on that interval. If $f'(x) = 0$ for every $x$ in an interval, then $f$ is constant on that interval.

**Definition 4** (Local extrema).

For a function $f(x)$ and a real number $c$ in its domain, a *local maximum* is a value $f(c)$ such that $f(c) \geq f(x)$ for all inputs $x$ near input $c$.

For a function $f(x)$ and a real number $c$ in its domain, a *local minimum* is a value $f(c)$ such that $f(c) \leq f(x)$ for all inputs $x$ near input $c$.

A *local extremum* is a local maximum or a local minimum of $f(x)$.

The plural is local extrema.

The value $f(0) = 0$ of $f(x) = x^3$ is not a local extremum, because the definition requires $f(0) \ge f(x)$ for every nearby input, or $f(0) \le f(x)$ for every nearby input. Every interval around $x = 0$ contains inputs $x$ with $x > 0$, where $f(x) = x^3 > 0$, and inputs $x$ with $x < 0$, where $f(x) = x^3 < 0$.

**Theorem 5** (Sign test for local extrema).

Let $f$ be continuous at an input $c$. If $f'$ changes from positive to negative at $c$, then $f(c)$ is a local maximum. If $f'$ changes from negative to positive at $c$, then $f(c)$ is a local minimum. If $f'$ has the same sign on both sides of $c$, then $f$ has no local extremum at $c$.

A function that is differentiable at an input is continuous there. The sign test therefore applies at every input $c$ where $f'(c)$ exists.

Each graph below shows $f$ near an input $c$, with the sign of $f'$ on each side.

**Local maximum**

![Figure](https://tangentspace.org/lessons/fig/e3a46d449cba.svg)

The graph rises to $(c, f(c))$ and then falls, and $f(c)$ is a local maximum.

**Local minimum**

![Figure](https://tangentspace.org/lessons/fig/c696c627d0dc.svg)

The graph falls to $(c, f(c))$ and then rises, and $f(c)$ is a local minimum.

**No local extremum**

![Figure](https://tangentspace.org/lessons/fig/901bdb21be2d.svg)

The graph rises on both sides of $c$, and $f(c)$ is neither a local maximum nor a local minimum.

*Breaks in the graph.* The sign test for local extrema applies only where $f$ is continuous. The function $h$ in Example 2 falls to the left of its jump at $x = 1$ and rises to the right. The value $h(1) = 2$ is neither a local maximum nor a local minimum, because the outputs just left of $x = 1$ are near $-1$ and the outputs just right of $x = 1$ are above $2$.

*Intervals and local extrema from $f'$.*

1. Find the inputs where $f'(x) = 0$ or where $f'(x)$ does not exist.
2. Evaluate $f'$ at one test input inside each interval those inputs mark off.
3. Label each interval increasing or decreasing by the sign of its test value.
4. At each input where $f$ is continuous, report a local maximum where $f'$ changes from positive to negative and a local minimum where $f'$ changes from negative to positive.

### Example 5

The derivative of $f(x) = x^3 - 5x$ is

$$f'(x) = 3x^2 - 5.$$

Identify where $f$ is increasing or decreasing, and identify its local extrema.

**Solution.**

Because the derivative $f'(x)$ exists at every input, $f$ is continuous everywhere, and the sign test for local extrema applies at every input. Find the inputs where $f'(x) = 0$ by setting $3x^2 - 5 = 0$:

$$x^2 = \frac{5}{3} \quad\text{so}\quad x = \pm\sqrt{5/3} \approx \pm 1.291.$$

Evaluate $f'$ at one test input in each of the three intervals those inputs mark off:

$$f'(-2) = 3(-2)^2 - 5 = 7, \qquad f'(0) = 3(0)^2 - 5 = -5, \qquad f'(2) = 3(2)^2 - 5 = 7.$$

Therefore, $f$ is increasing on $(-\infty, -\sqrt{5/3})$ and $(\sqrt{5/3}, \infty)$ and decreasing on $(-\sqrt{5/3}, \sqrt{5/3})$. Because the sign of $f'$ changes from positive to negative at $x = -\sqrt{5/3}$, $f$ has a local maximum there. By the definition of local extrema, the local maximum is the value $f\bigl(-\sqrt{5/3}\bigr)$:

$$f\bigl(-\sqrt{5/3}\bigr) = \bigl(-\sqrt{5/3}\bigr)^3 - 5\bigl(-\sqrt{5/3}\bigr)
= -\tfrac{5}{3}\sqrt{5/3} + 5\sqrt{5/3} = \tfrac{10}{3}\sqrt{5/3} \approx 4.303.$$

Because the sign of $f'$ changes from negative to positive at $x = \sqrt{5/3}$, $f$ has a local minimum there, and the local minimum is the value $f\bigl(\sqrt{5/3}\bigr)$:

$$f\bigl(\sqrt{5/3}\bigr) = \bigl(\sqrt{5/3}\bigr)^3 - 5\sqrt{5/3}
= \tfrac{5}{3}\sqrt{5/3} - 5\sqrt{5/3} = -\tfrac{10}{3}\sqrt{5/3} \approx -4.303.$$

On the graph of $f$, the local maximum, about 4.303, is the height of the peak at about $(-1.291, 4.303)$, and the local minimum, about $-4.303$, is the height of the valley at about $(1.291, -4.303)$.

![Figure](https://tangentspace.org/lessons/fig/728dede7424a.svg)

### Example 6

The derivative of a function $f(x)$ is

$$f'(x) = (x + 1)^2(x - 2).$$

Identify where $f$ is increasing or decreasing, and identify the inputs of its local extrema.

**Solution.**

The derivative is zero at $x = -1$ and $x = 2$, where the factors $(x + 1)^2$ and $x - 2$ are zero. Evaluate $f'$ at a test input in each interval, as in Example 5:

$$f'(-2) = (-1)^2(-4) = -4, \qquad f'(0) = (1)^2(-2) = -2, \qquad f'(3) = (4)^2(1) = 16.$$

Therefore, $f$ is decreasing on $(-\infty, -1)$ and $(-1, 2)$ and increasing on $(2, \infty)$. Because the derivative exists at every input, $f$ is continuous everywhere. The function $f$ has a local minimum at $x = 2$, because $f'$ changes from negative to positive there. The function has no local extremum at $x = -1$, because $f'$ is negative on both sides.

*A rising graph of $f'$ below the $x$-axis.* A graph of $f'$ that rises can be misread as a sign that $f$ is increasing. On $(0, \sqrt{5/3})$, the graph of $f'(x) = 3x^2 - 5$ from Example 5 rises while its values stay negative. The function $f$ is decreasing on $(0, \sqrt{5/3})$, because $f'(x) < 0$ for every $x$ in that interval.

## 4. Concavity and Points of Inflection

*Objective.* Given a first or second derivative, determine concavity and identify inflection points.

**Concave up**

![Figure](https://tangentspace.org/lessons/fig/ab8f76ff861a.svg)

The red segments are tangent lines. Where $f'' > 0$, their slopes $f'$ increase from left to right, and the graph bends upward.

**Concave down**

![Figure](https://tangentspace.org/lessons/fig/b495e9c7435d.svg)

The red segments are tangent lines. Where $f'' < 0$, their slopes $f'$ decrease from left to right, and the graph bends downward.

**Point of inflection**

![Figure](https://tangentspace.org/lessons/fig/e4153a07a5b1.svg)

The graph changes from concave down to concave up at $(c, f(c))$, where $f''$ changes sign.

**Definition 6** (Concavity).

For a function $f(x)$ differentiable on the interval $(a,b)$, $f(x)$ is *concave up* on $(a,b)$ if and only if $f'(x)$ is increasing on $(a,b)$.

For a function $f(x)$ differentiable on the interval $(a,b)$, $f(x)$ is *concave down* on $(a,b)$ if and only if $f'(x)$ is decreasing on $(a,b)$.

For a differentiable function, these definitions of concave up and concave down agree with the definitions in the [Lesson 5](https://tangentspace.org/lessons/5/) Notes.

On every interval, the line $f(x) = 2x + 1$ is neither concave up nor concave down, because the definitions require $f'$ to increase or to decrease and its derivative is the constant slope $f'(x) = 2$.

**Definition 7** (Point of inflection).

The point $(c, f(c))$ is a *point of inflection* of $f(x)$ if and only if $f(x)$ changes concavity at input $c$.

The point $(0, 0)$ on the graph of $y = x^2$ is not a point of inflection, because the definition requires a change of concavity and the graph is concave up on both sides of $x = 0$.

**Theorem 8** (Second derivative and concavity).

If $f''(x) > 0$ for every $x$ in an interval, then $f$ is concave up on that interval. If $f''(x) < 0$ for every $x$ in an interval, then $f$ is concave down on that interval. If $f''$ changes sign at an input $c$ and $f$ is continuous at $c$, then $(c, f(c))$ is a point of inflection of $f$.

The first two sentences of the theorem follow from the derivative and monotonicity theorem with $f'$ in place of $f$, together with the definition of concavity. Where $f'' > 0$, the derivative $f'$ is increasing, and by the definition of concavity, $f$ is concave up. Where $f'' < 0$, the derivative $f'$ is decreasing, and $f$ is concave down.

*Concavity and points of inflection from $f''$.*

1. Find the inputs where $f''(x) = 0$ or where $f''(x)$ does not exist.
2. Evaluate $f''$ at one test input inside each interval those inputs mark off.
3. Label each interval concave up or concave down by the sign of its test value.
4. Report a point of inflection $(c, f(c))$ at each input $c$ where the sign of $f''$ changes and $f$ is continuous.

### Example 7

The second derivative of $f(x) = x^3 - 5x$ is

$$f''(x) = 6x.$$

Determine the concavity of $f$ and identify its points of inflection.

**Solution.**

By the definition of concavity, $f$ is concave up on an interval where $f'$ increases, and $f'$ increases on each interval where $f'' > 0$. Likewise, $f$ is concave down on an interval where $f'$ decreases, and $f'$ decreases on each interval where $f'' < 0$. The second derivative $6x$ is zero at $x = 0$. Evaluate it at one test input on each side:

$$f''(-1) = 6(-1) = -6, \qquad f''(1) = 6(1) = 6.$$

Thus, $f$ is concave down on $(-\infty, 0)$ and concave up on $(0, \infty)$. The concavity changes at $x = 0$, and the output there is

$$f(0) = 0^3 - 5(0) = 0.$$

By the definition of a point of inflection, the point $(0, 0)$ is a point of inflection of $f$.

### Example 8

The second derivative of a function $f(x)$ satisfies

$$f''(x) > 0 \text{ on } (4, 6).$$

Determine whether each statement must be true:

1. The derivative $f'$ is increasing on $(4, 6)$.
2. The function $f$ is concave up on $(4, 6)$.
3. The function $f$ is increasing on $(4, 6)$.

**Solution.**

The first statement must be true by the derivative and monotonicity theorem applied to $f'$, because $f''(x) > 0$ on $(4, 6)$. The second statement must be true by the definition of concavity, because $f'$ is increasing on $(4, 6)$. The third statement need not be true, because whether $f$ is increasing on $(4, 6)$ depends on the sign of $f'$ there, and only the sign of $f''$ is given.

### Example 9

The second derivative of a function $f(x)$ is

$$f''(x) = (x - 1)^2.$$

Determine the concavity of $f$ and identify its points of inflection.

**Solution.**

The second derivative $(x - 1)^2$ is zero only at $x = 1$ and positive at every other input, because the square of a nonzero number is positive. The derivative $f'$ therefore increases on $(-\infty, 1)$ and on $(1, \infty)$. Because $f''(1)$ exists, $f'$ is continuous at $x = 1$ and increases across $x = 1$ as well. By the definition of concavity, $f$ is concave up on $(-\infty, \infty)$. The function $f$ has no point of inflection, because the concavity is the same on both sides of $x = 1$ and the definition of a point of inflection requires a change of concavity.

### Example 10

The graph below shows the derivative $f'(x)$ of a function $f(x)$ on $-3 \le x \le 5$. For $-3 < x < 5$, identify where $f$ is increasing or decreasing, the inputs of its local extrema, its concavity, and the inputs of its points of inflection.

![Figure](https://tangentspace.org/lessons/fig/1169b5524653.svg)

**Solution.**

The graph of $f'$ is below the $x$-axis on $(-3, -2)$ and $(1, 4)$ and above the $x$-axis on $(-2, 1)$ and $(4, 5)$. Therefore, $f$ is decreasing on $(-3, -2)$ and $(1, 4)$ and increasing on $(-2, 1)$ and $(4, 5)$. Because the derivative exists at every input of the graph, $f$ is continuous there. The function $f$ has local minima at $x = -2$ and $x = 4$, where the sign of $f'$ changes from negative to positive. The function $f$ has a local maximum at $x = 1$, where the sign changes from positive to negative.

On the graph, $f'$ has a local maximum at an input $x_1$ between $-1$ and 0 and a local minimum at an input $x_2$ between 2 and 3. The graph of $f'$ rises on $(-3, x_1)$ and $(x_2, 5)$ and falls on $(x_1, x_2)$. As a result, $f$ is concave up on $(-3, x_1)$ and $(x_2, 5)$ and concave down on $(x_1, x_2)$. The function $f$ has points of inflection at the inputs $x_1$ and $x_2$, because the concavity changes at each one.

The graph of $f'$ gives the inputs of the local extrema and the points of inflection of $f$. To find the values of the local extrema and the heights of the points of inflection, use a formula or a graph of $f$.

![Figure](https://tangentspace.org/lessons/fig/947726d18154.svg)