# IB Class Notes

## Second Derivative Test & Curve Sketching

Aims: By the end of this note, you will be able to

1. employ the first derivative to decide whether a function is increasing, decreasing or horizontal for a particular interval;
2. define a stationary point;
3. define an inflection point (or inflexion point), and
4. use the second derivative to find a maximum point or a minimum point.

### Increasing & Decreasing Functions

Derivatives can be used to help us to sketch curves- usually complicated ones. Here, we will start with some simple curves. In the process we will also learn about local maximum point, local minimum point and point of inflection.

Definition 1: Increasing Function
A function $f$ is said to be increasing in the interval between $x_1$ and $x_2$
if $x_1 > x_2$ then $f( x_1 ) > f( x_2 )$ (refer to Figure1).
We see that any tangent line drawn in the interval between $x_1$ and $x_2$ will have positive gradient.
That is $\frac{dy}{dx} = f '(x) > 0$ .

Definition 2: Decreasing Function
Similarly, a function $f$ is said to be decreasing in the interval between $x_3$ and $x_4$
if $x_4 > x_3$ then $f( x_4 ) < f( x_3)$ (refer to Figure1).
We see that any tangent line drawn in the interval between $x_3$ and $x_4$ will have negative gradient.
That is $\frac{dy}{dx} = f '(x) < 0$. In fact, studying Figure 1 closely shows that the function is decreasing for $x < a$ and $x > b$. The function is increasing for $a < x < b$ .

What happen to the function in Figure 1 at $x = a$ and $x = b$ ?

1. Points $( a, f(a) )$ and $( b, f(b) )$ are called turning points or stationary points. $( a, f(a) )$, for example, is a turning point because at this point, the gradient of a tangent to the curve changes from negative to positive. Stationary point is used and probably more appropriate and descriptive because $f '(a) = 0$; i.e. $\frac{dy}{dy} |_{x = a} = 0$ . That is, at this point the function is neither increasing nor decreasing.
2. The gradient of a stationary point is zero.
. $f '(a) = f '(b) = 0$.
3. Point $g$ in Figure 2 is also a stationary point and is called a stationary point of inflection. Notice that the immediate neighbourhood of point $g$ is flat.
4. Stationary points can be of three types:
1. local maximum
2. local minimum
3. point of inflection
5. A point of inflection can also be non-stationary. Such points are also called general points of inflection. Points $x, y, z,$ and $w$ in Figure 2 are general points of inflection. Notice, that these points unlike point $g$ have no immediate neighbourhoods that are flat.
6. Example 1
Find all stationary points of $f(x) = 6 + 9x - 3x^2 + x^3$ .
To find stationary point(s) we need to have $\frac{dy}{dx} = 0$ or $f '(x) = 0$. Thus, we need to first take the first derivative. $f '(x) = 9 - 6x + 3x^2$
Set $f '(x) = 0$ and solve for $x$ to obtain stationary point(s). \begin{align} 9 - 6x + 3x^2 & = 0 \\ (3x -3)(x-3) & = 0 \\ x = 1 ~ ~ \text{or} ~ ~ x & = 3 \end{align} To find the stationary points we need to find the corresponding $y$-coordinates.
$f(1) = 6 + 9 -3 +1 = 13$. So (1,13) is a stationary point.
$f(3) = 6 +27 -27 +27 = 33$. So (3,33) is also a stationary point.

Guided Example 2
Find all stationary points of $f(x) = x^2sin(x) + 1, ~ ~ 0 \leq x \leq 4$.
Note that when no unit is indicated then $x$ is conventionally taken to be in radian.

Find $f '(x) =$ .

Set $f '(x) = 0$ and solve for $x$.

$f '(x)=2xsinx + x^2cosx ;~ ~ x=0$ or $x \approx 2.29$ (3.s.f.)

7. Knowing that (1,13) and (3,33) in example 1 are stationary points is useful but often we will also like to know the nature of these points. For examples: Which of these is a local maximum?
Which of these is a local minimum?
Is (3,13) a point of inflection?
We will study two ways to determine the nature of a stationary point. These are Second Derivative Test
& First Derivative Test.
In this note, we will concentrate in the Second Derivative Test only. Another note will be devoted to First Derivative Test later.

### The Second Derivative Test.

 The second derivative of $y$ is represented by $\frac{d^2y}{dx^2}$ or $f''(x)$. Second derivative is just a derivative of the first derivative. Example 3 1. Take the first derivative of $y = 5x^2 + x^3$ with respect to $x$. $\frac{dy}{dx} = y ' (x ) = 10x + 3x^2$ . 2. Take the derivative of $y ' = 10x + 3x^2$ with respect to $x$. $\frac{dy'}{dx} = y '' (x ) = 10 + 6x$ . 3. The two instuctions above can be combined to read: Take the second derivative of $y = 5x^2 + x^3$ with respect to $x$. $y ' (x ) = 10x + 3x^2$. $\frac{d^2y}{dx^2} = y '' (x) = 10 + 6x$ We could also take the third derivative of $y = 5x^2 + x^3$. $y ''' (x) = y^{(3)} = 6$. $\frac{d^{n}y}{dx^{n}} = y^{(n)}(x)$ means the $n$th derivative of a function $y$. Note that the $n$th derivative is indicated by "(n)" where the bracket "( ) " is essential. The notation is incorrect if you replace the ( ) by either embrace { } or square bracket [ ] Example 4 \begin{align} y & = x sin x \\ y ' & = xcos x + sin x \\ y '' & = - xsin x + 2cos x \end{align} Example 5 \begin{align} y = \frac{x^5}{5} - x^3 + \frac{x}{2} \\ y ' = x^4 - 3x^2 + \frac{1}{2} \\ y '' = 4x^3 - 6 \end{align} Find the second derivatives of the following functions with respect to $x$: $y = sin x$ $y = \frac{1}{x}$ $y = (e^x)(cos x)$ Answers (a) -sin x (b) 2/x3 (c) -2exsinx

### Determining Local Maximum Point.

1. Figure 3 looks at a simple quadratic function with a maximum point. [In fact, in this case, ($x_1 , f( x_1 ))$ is the highest point on the curve. Such a point is sometimes known as the global maximum.] IB refers to such a point simply as the "turning point."
2. Notice that the first derivative of this function is a straight line with a negative slope. The line intercepts the $x$-axis at $x = x_1$, the maximum point.
3. Since the first derivative is a straight line with a negative slope than the second derivative in this case is simply a costant negative value.
4. For a more complicated function, the second derivative could be a function.
5. Example 6
Let $f(x) = -x^2 + 6x -2$
then $\frac{dy}{dx} = f '(x) = -2x + 6$
and $\frac{d^2y}{dx^2} = f ''(x) = -2$
Notice that the second derivative is negative.
6. Determining a Local Maximum
If $f'(a) = 0$ and $f '' (a) < 0$ then the stationary point $(a,f(a))$ is a local maximum point.

Concave Down
When a function has $f ''(x) < 0$ then the function $f$ is said to concaves down as in Figure 3.
Find the maximum point of the function $f$ in Figure 3.

7. Example 7
Find the local maximum point of $y= x^3- 6x^2 + 8$ .
Take the first derivative; $y '(x) = 3x^2-12x$
Setting $y '(x) = 0$ to find stationary point(s).
\begin{align} 3x^2-12x &= 0 \\ 3x ( x - 4 ) & = 0 \\ x = 0 or x &= 4 . \end{align}
Also, $y(0) = 8$ and $y (4) = -24$. Thus, the stationary points are (0, 8) and (4, -24). The next step is to determine which of these is a local maximum point.
$y ''(x) = 6x -12$ .
$y ''(0) < 0$ but $y ''(4) = 12 > 0$.
Thus, (0,8) is the local maximum point because its second derivative is negative.
We will see later that (4,-24) is a local minimum point.
8. Guided Example 8
Find the local maximum point in $y = (x^2)(x+3) - 9x + 2$.
Find $y '(x) =$
Set $y '(x) = 0$

$y ''(x) =$

### Determining Local Minimum Point

1. Similar method can be applied to determine local minimum point. Look at Figure 4 which depicts a simple quadratic function with minimum point. The first derivative is a straight line with positive slope and intercepts the $x$-axis at $x = x_1$, the minimum point. The second derivative in this case is a constant positive number because the first derivative is a straight line.

2. The second derivative can be a function if the original function is more complicated.
3. Thus, to determine whether a stationary point is a local minimum point or not, we will then simply check its second derivative.
4. Determining a Local Minimum
If $f'(a) = 0$ and $f '' (a) > 0$ then the stationary point $(a,f(a))$ is a local minimum point.

Concave Up
When a function has $f ''(x) > 0$ then the function $f$ is said to concaves up as in Figure 4.
Find the minimum point of the function $f$ in Figure 4.

5. Example 9
Find the minimum point in $y = x^2-5x + 12$.
Let us start by taking the first derivative.
$y ' (x) = 2x -5$
Set $y '(x) = 0$ to find stationary point(s). Thus, $2x -5 = 0$
$x = \frac{5}{2}$ and $y(\frac{5}{2}) = 18.25$
$y '' (x) = 2$ thus in this case, the second derivative is always positive.
The point (2.5, 18.25) is a minimum point.
6. Example 2
Find the local minimum in $f(x)= \frac{x^3}{3} - x^2 + 4$.
$f '(x) = x^2-2x$
Setting $f '(x) = 0$ to find stationary point(s). $x^2-2x = 0$
$x(x-2) = 0$
$x= 0$ or $x= 2$.
$f(x)=4$ and $f(2)= \frac{8}{3}$.
$f '' (x) = 2x -2$
$f ''(0) = -2 < 0$ and $f ''(2 )= 2 > 0$
So $(2, \frac{8}{3})$ is the local minimum point.

### What happen when $f ''(a) = 0$?

If $( a, f(a) )$ has $f '(x) = 0$ then $(a, f(a) )$ maybe a point of inflection. If $( a, f(a) )$ is a point of inflection then $f '(x) = 0$.
However, if $f '(x) = 0$ then this does not necessary imply the point $(a , f(a) )$ is a point of inflection.

A counter example
Consider $y = x^6$.
$y '(x)= 6x^5$
& $y ''(x) = 30 x^4$.
Let us set $y ''(x) = 0$ then we obtain $x = 0$ and $y(0) = 0$.
However, you can use your calculator to confirm that (0,0) is actually a minimum point and not a point of inflection.

We can either use graphical method or the first derivative test to determine whether or not a point is a point of inflection if $f '(a)=0$.
Alternatively, we can use the second derivative test as below.

Determining an Inflection point with Second Derivative Test
A point $(a , f(a) )$ is a point of inflection if and only if we have either
(i) $f ''(a - \delta) > 0$ and $f ''(a + \delta) < 0$
or
(ii) $f ''(a - \delta) < 0$ and $f ''(a + \delta) > 0$ where $\delta$ is a small real number.

In another word, $(a , f(a) )$ is a point of inflection if there is a change in concavity (i.e.a change in the signs of the second derivative) in the neigbourhood of $x = a$.
We will now use this new test to determine whether or not $y = x^6$ has a point of inflection at $x = 0$.
We have $y ''(x) = 30 x^4$ . Hence, we have $\begin{matrix} y''(0+0.1) = 30 (0.1)^4 ~ & ~ y ''(0 - 0.1) = 30 (- 0.1) ^4 \\ y''(0.1) > 0 ~ & ~ y''(-0.1) > 0 \end{matrix}$

We did not see any change in the signs of the second derivative in the calculations above and conclude that (0,0) in $y = x^6$ is NOT a point of inflection.

Exercise
Use the change in concavity concept to show that $y = (x-2)^3$ has a point of inflection in (2,0).

Email KokMing Lee