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 \( x \).

     

     

    Answers
    \( f '(x)=2xsinx + x^2cosx ;~ ~ x=0 \) 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 \(x\):

  1. \( y = sin x \)

  2. \( y = \frac{1}{x} \)

  3. \( y = (e^x)(cos x) \)

  4. 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).

 


First Derivative Test


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