Unit 4 - Differential Calculus
Required Prior Knowledge
Questions
If \(f\left(x\right)=x^{4}-3x^{2}+2x\) find the derivative \(f’\left(x\right)\).
Now find the derivative of the derivative.
Solutions
Get Ready
Questions
What does the derivative tell us about the graph of \(f\left(x\right)\)?
What does the derivative of the derivative tell us about the graph of \(f\left(x\right)\)?
Solutions
Notes
The second derivative is the derivative of the derivative.
It is denoted by \(f’’\left(x\right)\)or \(\frac{d^{2}y}{dx^{2}}\).
Your Turn
| \(f(x)\) | \(f'(x)\) | \(f''(x)\) |
|---|---|---|
| \(3x^2 - 6x + 2\) |
\(6x - 6\)
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\(6\)
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| \(3x - x^2 + 7\) |
\(3 - 2x\)
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\(-2\)
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| \(\frac{2}{\sqrt{x}} - 1\) |
\(-x^{-\frac{3}{2}}\)
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\(\frac{3}{2}x^{-\frac{5}{2}}\)
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| \(2x^3 - 3x^2 - x + 5\) |
\(6x^2 - 6x - 1\)
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\(12x - 6\)
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| \(\frac{2-3x}{x^2}\) |
\(-4x^{-3} + 3x^{-2}\)
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\(12x^{-4} - 6x^{-3}\)
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| \((1-2x)^2\) |
\(-4(1-2x)\)
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\(8\)
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Your Turn
| \(y\) | \(\frac{dy}{dx}\) | \(\frac{d^2y}{dx^2}\) |
|---|---|---|
| \(x - x^3\) |
\(1 - 3x^2\)
|
\(-6x\)
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| \(x^2 - \frac{5}{x^2}\) |
\(2x + 10x^{-3}\)
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\(2 - 30x^{-4}\)
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| \(2 - \frac{3}{\sqrt{x}}\) |
\(\frac{3}{2}x^{-\frac{3}{2}}\)
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\(-\frac{9}{4}x^{-\frac{5}{2}}\)
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| \(\frac{4-x}{x}\) |
\(-4x^{-2}\)
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\(8x^{-3}\)
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Your Turn
Find the value(s) of \(x\) such that \(f’\left(x\right)=0\) and \(f’’\left(x\right)=0\) for \(f\left(x\right)=2x^{3}-6x^{2}+6x+1\).
Notes
The second derivative describes the rate of change of the gradient.
This is known as the curvature or concavity of the graph of the function.
That is, it describes how ‘bendy’ the graph is at a given point.
When \(f’’\left(x\right)\gt 0\)
We say the function is concave up
Note that it doesn’t need to have a minimum
If \(f’\left(a\right)=0\) AND \(f’’\left(a\right)\gt 0\) then \(x=a\) is a local minimum
All tangents are below the curve
When \(f’’\left(x\right)\lt 0\)
We say the function is concave down
Note that it doesn’t need to have a maximum
If \(f’\left(a\right)=0\) AND \(f’’\left(a\right)\lt 0\) then \(x=a\) is a local maximum
All tangents are above the curve
Your Turn
For each point complete the table indicating whether each value is zero, positive or negative.
| Point | \(f(x)\) | \(f'(x)\) | \(f''(x)\) |
|---|---|---|---|
| A | 0 |
+
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-
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| B |
+
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0
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-
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| C |
-
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-
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0
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| D |
-
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0
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+
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| E |
+
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+ |
+
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| + | − | − | |
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-
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− | + |
Your Turn
For each point complete the table indicating whether each value is zero, positive or negative.
Are there any combinations that are NOT possible for this graph?
| Point | \(f(x)\) | \(f'(x)\) | \(f''(x)\) |
|---|---|---|---|
| A |
+
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+
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-
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| B |
0
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-
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+
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| C |
-
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0
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+
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| D |
+
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+
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0
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| E |
+
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0
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-
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| + | + | + | |
| − | − | - | |
| − | − | + |
Your Turn
From this graph determine when this function satisfies each condition in the table.
| Feature | Interval(s) |
|---|---|
| Positive |
\(x < -2 \\ 0 < x < 3\)
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| Negative |
\(-2 < x < 0 \\ x > 3\)
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| Increasing |
\(-1.1 < x < 1.8\)
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| Decreasing |
\(x < -1.1 \\ x > 1.8\)
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| Concave Up |
\(x < \frac{1}{3}\)
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| Concave Down |
\(x > \frac{1}{3}\)
?
|
Examples and Your Turns
Example
Determine when the following function is increasing and decreasing, and when it is concave up and concave down:$$y=2x^{3}-3x^{2}$$
Your Turn
Determine when the following function is increasing and decreasing, and when it is concave up and concave down:$$y=x^{2}+1$$
Your Turn
Determine when the following function is increasing and decreasing, and when it is concave up and concave down:$$y=\sqrt{x}-2$$
Your Turn
Determine when the following function is increasing and decreasing, and when it is concave up and concave down:$$y=x^{4}-12x^{2}$$
Your Turn
Determine when the following function is increasing and decreasing, and when it is concave up and concave down:$$y=x^{3}-3x^{2}+4x$$
Your Turn
What can you say about the function described in the table below? You can assume that all stationary points of the function are given. Sketch what the graph of the function might look like.
| Point | \(f(x)\) | \(f'(x)\) | \(f''(x)\) |
|---|---|---|---|
| \(x = -4\) | \(+\) | \(+\) | \(-\) |
| \(x = -3\) | \(+\) | \(0\) | \(-\) |
| \(x = -0.5\) | \(+\) | \(-\) | \(+\) |
| \(x = 0\) | \(0\) | \(0\) | \(0\) |
| \(x = 1\) | \(-\) | \(-\) | \(-\) |
| \(x = 2\) | \(-\) | \(0\) | \(+\) |
| \(x = 7\) | \(0\) | \(+\) | \(+\) |
Notes
The point when a graph changes from increasing to decreasing (or vice versa) is known as a turning point.
This is a special type of stationary point.
The point when a graph changes from concave up to concave down (or vice versa) is known as a Point of Inflection.
At a turning point \(\frac{dy}{dx}=0\)
At a point of inflection \(\frac{d^{2}y}{dx^{2}}=0\)
But consider the function \(f\left(x\right)=x^{4}\). Then$$f’\left(x\right)=4x^{3}\\f’’\left(x\right)=12x^{2}$$So the point where \(f’’\left(x\right)=0\) is \(x=0\).
But from the graph we know this is NOT a point of inflection.
There are stationary points that are not turning points i.e. where \(\frac{dy}{dx}=0\) but the sign does not change.
Similarly, there are points with \(\frac{d^{2}y}{dx^{2}}=0\) that are not points of inflection.
To find a point of inflection we must have BOTH
\(f’’\left(x\right)=0\) AND
A change in sign of \(f’’\left(x\right)\)
Examples and Your Turns
Example
Consider the function \(y=x^{4}\left(5-x\right)\). Find the points of inflection.
Example
Consider the function \(f\left(x\right)=2x^{4}-4x^{2}+1\).
Find and classify the stationary points.
Find and classify the points of inflection.
Determine when the function is increasing and decreasing.
Determine when the function is concave up and concave down.
Sketch the graph of \(y=f\left(x\right)\).
1. Derivatives
\begin{align}
f'(x) &= 8x^3 - 8x \\
&= 8x(x^2 - 1) \\
&= 8x(x - 1)(x + 1) \\[10pt]
f''(x) &= 24x^2 - 8 \\
&= 8(3x^2 - 1)
\end{align}
2. Stationary Points
Find and classify the stationary points:
Set \(f'(x) = 0\):
$$8x(x - 1)(x + 1) = 0 \implies x = 0, \; x = 1, \; x = -1$$
Classification using the Second Derivative Test:
- At \(x = 0\): \(f(0) = 1 \implies (0,1)\) is a stationary point $$f''(0) = -8 < 0 \implies \text{concave down} \implies \text{local maximum}$$
- At \(x = 1\): \(f(1) = -1 \implies (1,-1)\) is a stationary point $$f''(1) = 16 > 0 \implies \text{concave up} \implies \text{local minimum}$$
- At \(x = -1\): \(f(-1) = -1 \implies (-1,-1)\) is a stationary point $$f''(-1) = 16 > 0 \implies \text{concave up} \implies \text{local minimum}$$
3. Intervals of Increase & Decrease
Determine when the function is increasing and decreasing:
Testing monotonic intervals around the stationary points \(x = -1\), \(x = 0\), and \(x = 1\):
| \(x\) | \(-2\) | \(-1\) | \(-0.5\) | \(0\) | \(0.5\) | \(1\) | \(2\) |
|---|---|---|---|---|---|---|---|
| \(f'(x)\) | \(-\) | \(0\) | \(+\) | \(0\) | \(-\) | \(0\) | \(+\) |
- Increasing (\(f'(x) > 0\)): $$-1 < x < 0 \quad \text{and} \quad x > 1$$
- Decreasing (\(f'(x) < 0\)): $$x < -1 \quad \text{and} \quad 0 < x < 1$$
4. Points of Inflection & Concavity
Find and classify the points of inflection:
Set \(f''(x) = 0\):
$$8(3x^2 - 1) = 0 \implies x = \pm\frac{1}{\sqrt{3}}$$
Testing concavity intervals around \(x = \pm\frac{1}{\sqrt{3}}\):
| \(x\) | \(-2\) | \(-\frac{1}{\sqrt{3}}\) | \(0\) | \(\frac{1}{\sqrt{3}}\) | \(2\) |
|---|---|---|---|---|---|
| \(f''(x)\) | \(+\) | \(0\) | \(-\) | \(0\) | \(+\) |
Determine when the function is concave up and concave down:
- Concave Up (\(f''(x) > 0\)): $$x < -\frac{1}{\sqrt{3}} \quad \text{and} \quad x > \frac{1}{\sqrt{3}}$$
- Concave Down (\(f''(x) < 0\)): $$-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$$
5. Graph
Your Turn
Consider the function \(f\left(x\right)=3x^{4}+4x^{3}-2\).
Find and classify the stationary points.
Find and classify the points of inflection.
Determine when the function is increasing and decreasing.
Determine when the function is concave up and concave down.
Sketch the graph of \(y=f\left(x\right)\).
1. Derivatives
\begin{align}
f'(x) &= 12x^3 + 12x^2 \\
&= 12x^2(x + 1) \\[10pt]
f''(x) &= 36x^2 + 24x \\
&= 12x(3x + 2)
\end{align}
2. Stationary Points
Find and classify the stationary points:
Set \(f'(x) = 0\):
$$12x^2(x + 1) = 0 \implies x = 0, \; x = -1$$
Classification using the Second Derivative Test:
- At \(x = -1\): \(f(-1) = -3 \implies (-1,-3)\) is a stationary point $$f''(-1) = 12 > 0 \implies \text{concave up} \implies \text{local minimum}$$
- At \(x = 0\): \(f(0) = -2 \implies (0,-2)\) is a stationary point $$f''(0) = 0 \implies \text{test fails (inflection point layout below)}$$
3. Intervals of Increase & Decrease
Determine when the function is increasing and decreasing:
Testing monotonic intervals around the stationary points \(x = -1\) and \(x = 0\):
| \(x\) | \(-2\) | \(-1\) | \(-0.5\) | \(0\) | \(1\) |
|---|---|---|---|---|---|
| \(f'(x)\) | \(-\) | \(0\) | \(+\) | \(0\) | \(+\) |
- Increasing (\(f'(x) > 0\)): $$-1 < x < 0 \quad \text{and} \quad x > 0 \quad (\text{or simply } x > -1 \text{ except at } x=0)$$
- Decreasing (\(f'(x) < 0\)): $$x < -1$$
4. Points of Inflection & Concavity
Find and classify the points of inflection:
Set \(f''(x) = 0\):
$$12x(3x + 2) = 0 \implies x = 0, \; x = -\frac{2}{3}$$
Testing concavity intervals around \(x = -\frac{2}{3}\) and \(x = 0\):
| \(x\) | \(-1\) | \(-\frac{2}{3}\) | \(-0.5\) | \(0\) | \(1\) |
|---|---|---|---|---|---|
| \(f''(x)\) | \(+\) | \(0\) | \(-\) | \(0\) | \(+\) |
Determine when the function is concave up and concave down:
- Concave Up (\(f''(x) > 0\)): $$x < -\frac{2}{3} \quad \text{and} \quad x > 0$$
- Concave Down (\(f''(x) < 0\)): $$-\frac{2}{3} < x < 0$$
5. Graph
Your Turn
Consider the function \(f\left(x\right)=2x^{3}+6x^{2}+1\).
Find and classify the stationary points.
Find and classify the points of inflection.
Determine when the function is increasing and decreasing.
Determine when the function is concave up and concave down.
Sketch the graph of \(y=f\left(x\right)\).
1. Derivatives
\begin{align}
f'(x) &= 6x^2 + 12x \\
&= 6x(x + 2) \\[10pt]
f''(x) &= 12x + 12 \\
&= 12(x + 1)
\end{align}
2. Stationary Points
Find and classify the stationary points:
Set \(f'(x) = 0\):
$$6x(x + 2) = 0 \implies x = 0, \; x = -2$$
Classification using the Second Derivative Test:
- At \(x = -2\): \(f(-2) = 9 \implies (-2,9)\) is a stationary point $$f''(-2) = -12 < 0 \implies \text{concave down} \implies \text{local maximum}$$
- At \(x = 0\): \(f(0) = 1 \implies (0,1)\) is a stationary point $$f''(0) = 12 > 0 \implies \text{concave up} \implies \text{local minimum}$$
3. Intervals of Increase & Decrease
Determine when the function is increasing and decreasing:
Testing monotonic intervals around the stationary points \(x = -2\) and \(x = 0\):
| \(x\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) |
|---|---|---|---|---|---|
| \(f'(x)\) | \(+\) | \(0\) | \(-\) | \(0\) | \(+\) |
- Increasing (\(f'(x) > 0\)): $$x < -2 \quad \text{and} \quad x > 0$$
- Decreasing (\(f'(x) < 0\)): $$-2 < x < 0$$
4. Points of Inflection & Concavity
Find and classify the points of inflection:
Set \(f''(x) = 0\):
$$12(x + 1) = 0 \implies x = -1$$
Testing concavity intervals around \(x = -1\):
| \(x\) | \(-2\) | \(-1\) | \(0\) |
|---|---|---|---|
| \(f''(x)\) | \(-\) | \(0\) | \(+\) |
Determine when the function is concave up and concave down:
- Concave Up (\(f''(x) > 0\)): $$x > -1$$
- Concave Down (\(f''(x) < 0\)): $$x < -1$$
5. Graph
Your Turn
Consider the function \(f\left(x\right)=3x^{4}-16x^{3}+24x^{2}-9\).
Find and classify the stationary points.
Find and classify the points of inflection.
Determine when the function is increasing and decreasing.
Determine when the function is concave up and concave down.
Sketch the graph of \(y=f\left(x\right)\).
1. Derivatives
\begin{align}
f'(x) &= 12x^3 - 48x^2 + 48x \\
&= 12x(x^2 - 4x + 4) \\
&= 12x(x - 2)^2 \\[10pt]
f''(x) &= 36x^2 - 96x + 48 \\
&= 12(3x^2 - 8x + 4) \\
&= 12(3x - 2)(x - 2)
\end{align}
2. Stationary Points
Find and classify the stationary points:
Set \(f'(x) = 0\):
$$12x(x - 2)^2 = 0 \implies x = 0, \; x = 2$$
Classification using the Second Derivative Test:
- At \(x = 0\): \(f(0) = -9 \implies (0,-9)\) is a stationary point $$f''(0) = 48 > 0 \implies \text{concave up} \implies \text{local minimum}$$
- At \(x = 2\): \(f(2) = 7 \implies (2,7)\) is a stationary point $$f''(2) = 0 \implies \text{test fails (inflection point layout below)}$$
3. Intervals of Increase & Decrease
Determine when the function is increasing and decreasing:
Testing monotonic intervals around the stationary points \(x = 0\) and \(x = 2\):
| \(x\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|---|
| \(f'(x)\) | \(-\) | \(0\) | \(+\) | \(0\) | \(+\) |
- Increasing (\(f'(x) > 0\)): $$0 < x < 2 \quad \text{and} \quad x > 2 \quad (\text{or simply } x > 0 \text{ except at } x=2)$$
- Decreasing (\(f'(x) < 0\)): $$x < 0$$
4. Points of Inflection & Concavity
Find and classify the points of inflection:
Set \(f''(x) = 0\):
$$12(3x - 2)(x - 2) = 0 \implies x = \frac{2}{3}, \; x = 2$$
Testing concavity intervals around \(x = \frac{2}{3}\) and \(x = 2\):
| \(x\) | \(0\) | \(\frac{2}{3}\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|---|
| \(f''(x)\) | \(+\) | \(0\) | \(-\) | \(0\) | \(+\) |
Determine when the function is concave up and concave down:
- Concave Up (\(f''(x) > 0\)): $$x < \frac{2}{3} \quad \text{and} \quad x > 2$$
- Concave Down (\(f''(x) < 0\)): $$\frac{2}{3} < x < 2$$
5. Graph
Your Turn
Consider the function \(f\left(x\right)=2x^{3}-9x^{2}+12x\).
Find and classify the stationary points.
Find and classify the points of inflection.
Determine when the function is increasing and decreasing.
Determine when the function is concave up and concave down.
Sketch the graph of \(y=f\left(x\right)\).
1. Derivatives
\begin{align}
f'(x) &= 6x^2 - 18x + 12 \\
&= 6(x^2 - 3x + 2) \\
&= 6(x - 1)(x - 2) \\[10pt]
f''(x) &= 12x - 18 \\
&= 6(2x - 3)
\end{align}
2. Stationary Points
Find and classify the stationary points:
Set \(f'(x) = 0\):
$$6(x - 1)(x - 2) = 0 \implies x = 1, \; x = 2$$
Classification using the Second Derivative Test:
- At \(x = 1\): \(f(1) = 5 \implies (1,5)\) is a stationary point $$f''(1) = -6 < 0 \implies \text{concave down} \implies \text{local maximum}$$
- At \(x = 2\): \(f(2) = 4 \implies (2,4)\) is a stationary point $$f''(2) = 6 > 0 \implies \text{concave up} \implies \text{local minimum}$$
3. Intervals of Increase & Decrease
Determine when the function is increasing and decreasing:
Testing monotonic intervals around the stationary points \(x = 1\) and \(x = 2\):
| \(x\) | \(0\) | \(1\) | \(1.5\) | \(2\) | \(3\) |
|---|---|---|---|---|---|
| \(f'(x)\) | \(+\) | \(0\) | \(-\) | \(0\) | \(+\) |
- Increasing (\(f'(x) > 0\)): $$x < 1 \quad \text{and} \quad x > 2$$
- Decreasing (\(f'(x) < 0\)): $$1 < x < 2$$
4. Points of Inflection & Concavity
Find and classify the points of inflection:
Set \(f''(x) = 0\):
$$6(2x - 3) = 0 \implies x = \frac{3}{2}$$
Testing concavity intervals around \(x = \frac{3}{2}\):
| \(x\) | \(1\) | \(\frac{3}{2}\) | \(2\) |
|---|---|---|---|
| \(f''(x)\) | \(-\) | \(0\) | \(+\) |
Determine when the function is concave up and concave down:
- Concave Up (\(f''(x) > 0\)): $$x > \frac{3}{2}$$
- Concave Down (\(f''(x) < 0\)): $$x < \frac{3}{2}$$
5. Graph
Your Turn
Consider the function \(f\left(x\right)=x\left(x-3\right)^{2}\).
Find and classify the stationary points.
Find and classify the points of inflection.
Determine when the function is increasing and decreasing.
Determine when the function is concave up and concave down.
Sketch the graph of \(y=f\left(x\right)\).
1. Derivatives
\begin{align}
f(x) &= x^3 - 6x^2 + 9x \\[10pt]
f'(x) &= 3x^2 - 12x + 9 \\
&= 3(x^2 - 4x + 3) \\
&= 3(x - 1)(x - 3) \\[10pt]
f''(x) &= 6x - 12 \\
&= 6(x - 2)
\end{align}
2. Stationary Points
Find and classify the stationary points:
Set \(f'(x) = 0\):
$$3(x - 1)(x - 3) = 0 \implies x = 1, \; x = 3$$
Classification using the Second Derivative Test:
- At \(x = 1\): \(f(1) = 4 \implies (1,4)\) is a stationary point $$f''(1) = -6 < 0 \implies \text{concave down} \implies \text{local maximum}$$
- At \(x = 3\): \(f(3) = 0 \implies (3,0)\) is a stationary point $$f''(3) = 6 > 0 \implies \text{concave up} \implies \text{local minimum}$$
3. Intervals of Increase & Decrease
Determine when the function is increasing and decreasing:
Testing monotonic intervals around the stationary points \(x = 1\) and \(x = 3\):
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|
| \(f'(x)\) | \(+\) | \(0\) | \(-\) | \(0\) | \(+\) |
- Increasing (\(f'(x) > 0\)): $$x < 1 \quad \text{and} \quad x > 3$$
- Decreasing (\(f'(x) < 0\)): $$1 < x < 3$$
4. Points of Inflection & Concavity
Find and classify the points of inflection:
Set \(f''(x) = 0\):
$$6(x - 2) = 0 \implies x = 2$$
Testing concavity intervals around \(x = 2\):
| \(x\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|
| \(f''(x)\) | \(-\) | \(0\) | \(+\) |
Determine when the function is concave up and concave down:
- Concave Up (\(f''(x) > 0\)): $$x > 2$$
- Concave Down (\(f''(x) < 0\)): $$x < 2$$
5. Graph
Your Turn
Consider the function \(f\left(x\right)=e^{-x^{2}}\).
Find and classify the stationary points.
Find and classify the points of inflection.
Determine when the function is increasing and decreasing.
Determine when the function is concave up and concave down.
Sketch the graph of \(y=f\left(x\right)\).
1. Derivatives
\begin{align}
f(x) &= e^{-x^2} \\[10pt]
f'(x) &= -2x e^{-x^2} \\[10pt]
f''(x) &= -2e^{-x^2} + (-2x)(-2x)e^{-x^2} \\
&= -2e^{-x^2} + 4x^2e^{-x^2} \\
&= 2e^{-x^2}(2x^2 - 1)
\end{align}
2. Stationary Points
Find and classify the stationary points:
Set \(f'(x) = 0\):
$$-2x e^{-x^2} = 0$$
Since \(e^{-x^2} \neq 0\) for all real \(x\):
$$-2x = 0 \implies x = 0$$
Classification using the Second Derivative Test:
- At \(x = 0\): \(f(0) = 1 \implies (0,1)\) is a stationary point $$f''(0) = 2(1)(-1) = -2 < 0 \implies \text{concave down} \implies \text{local maximum}$$
3. Intervals of Increase & Decrease
Determine when the function is increasing and decreasing:
Testing monotonic intervals around the stationary point \(x = 0\):
| \(x\) | \(-1\) | \(0\) | \(1\) |
|---|---|---|---|
| \(f'(x)\) | \(+\) | \(0\) | \(-\) |
- Increasing (\(f'(x) > 0\)): $$x < 0$$
- Decreasing (\(f'(x) < 0\)): $$x > 0$$
4. Points of Inflection & Concavity
Find and classify the points of inflection:
Set \(f''(x) = 0\):
$$2e^{-x^2}(2x^2 - 1) = 0 \implies 2x^2 - 1 = 0 \implies x = \pm\frac{1}{\sqrt{2}}$$
Testing concavity intervals around \(x = -\frac{1}{\sqrt{2}}\) and \(x = \frac{1}{\sqrt{2}}\):
| \(x\) | \(-1\) | \(-\frac{1}{\sqrt{2}}\) | \(0\) | \(\frac{1}{\sqrt{2}}\) | \(1\) |
|---|---|---|---|---|---|
| \(f''(x)\) | \(+\) | \(0\) | \(-\) | \(0\) | \(+\) |
Determine when the function is concave up and concave down:
- Concave Up (\(f''(x) > 0\)): $$x < -\frac{1}{\sqrt{2}} \quad \text{and} \quad x > \frac{1}{\sqrt{2}}$$
- Concave Down (\(f''(x) < 0\)): $$-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$$
5. Graph
Key Facts
Use this applet to generate a prompt for a Key Fact that you need to know for the course. The idea is that you should KNOW these key facts in order to be able to solve problems.
Taking it Deeper
Conceptual Questions to Consider
Why do we have to check for a change in sign of \(f’’\left(x\right)\) to confirm a point of inflection?
Can a linear function have a point of inflection? What about a quadratic? Why?
Which other function types never have a point of inflection?
How can you use the second derivative and concavity to test if a stationary point is a maximum or minimum? Why does this not work if \(f’’\left(x\right)=0\)?
What is the difference between a function being positive/negative, increasing/decreasing and concave up/down?
Common Mistakes / Misconceptions
The biggest misconception is assuming that whenever \(f’’\left(x\right)=0\) there is a point of inflection.
Using either \(f’\left(x\right)\) or \(f’’\left(x\right)\) to try to calculate the \(y\) value once the \(x\) coordinate is known.
Connecting This to Other Skills
This skill builds on everything we have already seen in Differential Calculus (Unit 4) and is particularly closely related to the idea of Stationary Points (4.14).
The second derivative is a stepping stone to the ideas of Higher Derivatives (4.16), and is very helpful for Graphing Derivatives (4.17).
The connection between displacement and acceleration uses the second derivative in Kinematics (5.15).
Self-Reflection
What was the most challenging part of this skill for you?
What are you still unsure about that you need to review?