Simple Applications of Derivatives
Learn to use derivative signs to analyze functions, classify extrema, solve basic optimization problems, and interpret motion.
1. Why are derivatives useful?
A derivative measures how a function changes.
In the previous lesson, we interpreted:
\[f'(x)\]
as the slope of the tangent to the graph of
\(f\)
We can use these slopes to answer questions such as:
- Where does a function increase or decrease?
- Where does it have a local maximum or minimum?
- What dimensions produce the largest possible area?
- When is a moving object traveling in the positive or negative direction?
The main strategy is:
2. Using derivatives to determine monotonicity
Suppose
\(f\)
is differentiable throughout an open interval. If:
\[f'(x)>0\]
throughout that interval, then
\(f\)
is strictly increasing there. If:
\[f'(x)<0\]
throughout that interval, then
\(f\)
is strictly decreasing there. If:
\[f'(x)=0\]
at every point of the interval, then
\(f\)
is constant there.
Important distinction
A derivative equal to zero at one point does not mean the function is constant. For example:
\[f(x)=x^3\]
has:
\[f'(x)=3x^2,\]
so:
\[f'(0)=0.\]
Nevertheless, the function is strictly increasing on all real numbers. Strict increase can occur even when the derivative is zero at isolated points.
| Derivative condition | Function behavior |
|---|---|
| \(f'(x)>0\) throughout the interval | Strictly increasing |
| \(f'(x)<0\) throughout the interval | Strictly decreasing |
| \(f'(x)=0\) throughout the interval | Constant |
Worked example 1: A linear function
Determine the monotonicity of:
\[f(x)=5x-7.\]
Differentiate:
\[f'(x)=5.\]
Since:
\[5>0,\]
the function is strictly increasing on:
\[(-\infty,\infty).\]
Worked example 2: A quadratic function
Find where:
\[f(x)=x^2-4x+3\]
increases and decreases.
Differentiate:
\[f'(x)=2x-4=2(x-2).\]
For:
\[x<2,\]
we have:
\[f'(x)<0.\]
For:
\[x>2,\]
we have:
\[f'(x)>0.\]
Therefore:
\[\text{Decreasing on }(-\infty,2),\]
\[\text{Increasing on }(2,\infty).\]
In this lesson, sign-analysis intervals are written as open intervals. For continuous examples such as this quadratic, the turning-point endpoint can also be included in each adjacent monotonic interval.
3. Stationary points and critical points
3.1 Stationary points
A stationary point occurs at an input
\(c\)
where:
\[f'(c)=0.\]
The graph has a horizontal tangent there. The coordinates are:
\[(c,f(c)).\]
3.2 Critical numbers and critical points
In this lesson, an interior input
\(c\)
in the domain is called a critical number if:
\[f'(c)=0\]
or the derivative does not exist there. The corresponding graph point:
\[(c,f(c))\]
is a critical point. A stationary point is therefore one type of critical point.
Example: A critical point without a derivative
Consider:
\[f(x)=|x|.\]
At:
\[x=0,\]
the graph has a sharp corner and no derivative. However,
\(0\)
belongs to the domain, so it is a critical number.
Domain warning
An input where the function itself is undefined is not a critical number. For example:
\[f(x)=\frac1x\]
is undefined at
\(x=0\)
Therefore,
\(0\)
is not a critical number, although it separates the domain into two intervals.
Why do we examine critical points?
- If a function has a local maximum or minimum at an interior point and is differentiable there, then its derivative must be zero.
- Therefore, critical points are candidates for local extrema.
- They are not automatically maxima or minima.
4. Local maximum and minimum values
Local maximum
A function has a local maximum at
\(x=c\)
if its value there is at least as large as nearby function values.
Local minimum
A function has a local minimum at
\(x=c\)
- if its value there is at most as large as nearby function values. The word local means that we compare with nearby points.
- An absolute maximum is the largest value over the entire domain being considered.
- An absolute minimum is the smallest value over that domain.
Distinguish three answers
Suppose a function has a local minimum at the point:
\[(2,-5).\]
Then:
- The minimum occurs at the input \(x=2\).
- The local minimum value is \(-5\).
- The local minimum point is \((2,-5)\).
Answer the quantity requested in the question.
5. The first-derivative sign test
To classify a critical point, inspect the derivative immediately on both sides.
Positive to negative
If the derivative changes from positive to negative, the function increases and then decreases. Therefore, the point is a local maximum.
Negative to positive
If the derivative changes from negative to positive, the function decreases and then increases. Therefore, the point is a local minimum.
No sign change
If the derivative keeps the same strict sign on both sides, the function continues in the same direction. The critical point is neither a local maximum nor a local minimum. These tests apply when the function is continuous at the point and has the stated derivative signs on nearby intervals.
| Sign before | Sign after | Conclusion |
|---|---|---|
| Positive | Negative | Local maximum |
| Negative | Positive | Local minimum |
| Positive | Positive | Neither |
| Negative | Negative | Neither |
6. Derivative sign analysis: A complete example
Worked example 3
Analyze:
\[f(x)=x^3-3x^2+2.\]
Find:
- Stationary points.
- Increasing and decreasing intervals.
- Local maximum and minimum values.
Step 1: Check the domain
This is a polynomial, so:
\[\text{Domain}=(-\infty,\infty).\]
Its derivative exists everywhere.
Step 2: Differentiate
\[f'(x)=3x^2-6x.\]
Factor:
\[f'(x)=3x(x-2).\]
Step 3: Find stationary inputs
Solve:
\[3x(x-2)=0.\]
Therefore:
\[x=0 \quad\text{or}\quad x=2.\]
Step 4: Divide the number line
The stationary inputs separate the domain into:
\[(-\infty,0), \qquad (0,2), \qquad (2,\infty).\]
Step 5: Determine the derivative sign
For
\(x<0\)
both
\(x\)
and
\(x-2\)
are negative, so their product is positive.
For
\(0<x<2\)
the factors have opposite signs, so the product is negative.
For
\(x>2\)
both factors are positive.
| Interval | Sign of \(x\) | Sign of \(x-2\) | Sign of \(f'(x)\) | Behavior of \(f\) |
|---|---|---|---|---|
| \((-\infty,0)\) | Negative | Negative | Positive | Increasing |
| \((0,2)\) | Positive | Negative | Negative | Decreasing |
| \((2,\infty)\) | Positive | Positive | Positive | Increasing |
Step 6: Classify the stationary points
At
\(x=0\)
the derivative changes from positive to negative. Therefore, there is a local maximum. Calculate:
\[f(0)=2.\]
The local maximum point is:
\[(0,2).\]
At
\(x=2\)
the derivative changes from negative to positive. Therefore, there is a local minimum. Calculate:
\[f(2)=8-12+2=-2.\]
The local minimum point is:
\[(2,-2).\]
Final results
The function increases on:
\[(-\infty,0)
\quad\text{and}\quad
(2,\infty).\]
It decreases on:
\[(0,2).\]
Its local maximum value is:
\[2.\]
Its local minimum value is:
\[-2.\]
These are local extrema. The cubic has no absolute maximum or minimum on all real numbers.
7. A stationary point with no maximum or minimum
Worked example 4
Analyze:
\[g(x)=(x-1)^3.\]
Expand if needed:
\[g(x)=x^3-3x^2+3x-1.\]
Differentiate:
\[g'(x)=3x^2-6x+3.\]
Factor:
\[g'(x)=3(x-1)^2.\]
The derivative is zero at:
\[x=1.\]
For every other input:
\[3(x-1)^2>0.\]
Thus, the derivative is positive on both sides of
\(1\)
The function continues increasing through the stationary point:
\[(1,0).\]
This point is neither a local maximum nor a local minimum.
Conclusion:
Solving
\(f'(x)=0\)
identifies candidates. The sign test determines their behavior.
8. Absolute maximum and minimum on a closed interval
For a continuous function on a closed, bounded interval:
\[[a,b],\]
an absolute maximum and an absolute minimum exist.
To find them:
- Find the interior critical numbers.
- Keep only those inside the interval.
- Evaluate the function at those inputs.
- Evaluate the function at both endpoints.
- Compare all values.
Endpoints must be checked even when their derivatives are not zero.
Worked example 5
Find the absolute maximum and minimum of:
\[f(x)=x^2-4x+1\]
on:
\[[0,5].\]
Differentiate:
\[f'(x)=2x-4.\]
The interior stationary input is:
\[x=2.\]
Evaluate the candidates:
\[f(0)=1,\]
\[f(2)=4-8+1=-3,\]
\[f(5)=25-20+1=6.\]
Therefore, the absolute minimum is:
\[-3
\quad\text{at}\quad
x=2.\]
The absolute maximum is:
\[6
\quad\text{at}\quad
x=5.\]
The maximum occurs at an endpoint, where the derivative is not zero.
9. Simple optimization problems
An optimization problem asks for the largest or smallest possible value of a quantity.
Use this method:
- Define the variable.
- Write the quantity to optimize as a function.
- State the feasible domain.
- Differentiate.
- Find and check candidate inputs.
- Interpret the result with units.
Worked example 6: Maximum rectangle area
A rectangle has perimeter
\(40\)
meters. Find the dimensions giving the largest area.
Step 1: Define the variables
Let the sides be:
\[x\text{ meters}
\quad\text{and}\quad
y\text{ meters}.\]
The perimeter condition is:
\[2x+2y=40.\]
Therefore:
\[y=20-x.\]
Step 2: Write the area function
\[A=xy.\]
Substitute:
\[A(x)=x(20-x).\]
\[A(x)=20x-x^2.\]
Step 3: State the domain
Both side lengths must be positive:
\[0<x<20.\]
Step 4: Differentiate
\[A'(x)=20-2x.\]
Set equal to zero:
\[20-2x=0.\]
\[x=10.\]
Step 5: Check the sign change
For
\(x<10\)
\[A'(x)>0.\]
For
\(x>10\)
\[A'(x)<0.\]
The area increases and then decreases throughout the feasible domain. Therefore,
\(x=10\)
gives the absolute maximum.
Step 6: Interpret
\[y=20-10=10.\]
The optimal dimensions are:
\[10\text{ m}\times10\text{ m}.\]
The maximum area is:
\[A(10)=100\text{ m}^2.\]
Worked example 7: Minimum of a modeled cost
A simplified cost model is:
\[C(x)=x^2-12x+50,
\qquad
0\leq x\leq10.\]
Find its minimum.
Differentiate:
\[C'(x)=2x-12.\]
Solve:
\[2x-12=0.\]
\[x=6.\]
Compare the interior candidate and endpoints:
\[C(0)=50,\]
\[C(6)=36-72+50=14,\]
\[C(10)=100-120+50=30.\]
The minimum modeled cost is:
\[14\]
at:
\[x=6.\]
If
\(x\)
represents a whole-number quantity, feasible integer inputs must be considered. Here,
\(6\)
is already an integer. A non-integer stationary input would require checking nearby feasible integers.
10. Motion applications: Position, displacement, and velocity
Suppose an object moves along a straight line.
Let:
\[s(t)\]
represent its position at time
\(t\)
Position may be positive or negative depending on the chosen origin and positive direction.
10.1 Displacement
The displacement from time
\(t_1\)
to time
\(t_2\)
is:
\[s(t_2)-s(t_1).\]
Displacement records the net change in position.
10.2 Average velocity
\[v_{\text{average}}
=
\frac{s(t_2)-s(t_1)}{t_2-t_1}.\]
This is the slope of a secant on a position-time graph.
10.3 Instantaneous velocity
Velocity at a particular time is:
\[v(t)=s'(t).\]
This is the slope of the tangent to the position-time graph. If position is measured in meters and time in seconds, velocity is measured in:
\[\text{m/s}.\]
10.4 Direction of motion
- If \(v(t)>0\), the object moves in the positive direction.
- If \(v(t)<0\), it moves in the negative direction.
- If \(v(t)=0\), it is instantaneously at rest.
A zero velocity does not automatically mean a change of direction. Check the signs on both sides.
10.5 Speed
Speed is the magnitude of velocity:
\[\text{Speed}=|v(t)|.\]
Speed is never negative.
Worked example 8: An object that reverses direction
An object’s position is:
\[s(t)=t^2-6t+8,
\qquad
0\leq t\leq6.\]
Position is measured in meters and time in seconds.
Find the velocity function.
\[v(t)=s'(t)=2t-6.\]
When is the object at rest?
\[2t-6=0.\]
\[t=3.\]
Determine its direction.
For:
\[0<t<3,\]
velocity is negative.
For:
\[3<t<6,\]
velocity is positive. The object reverses direction at
\(t=3\)
Find its position when it reverses.
\[s(3)=9-18+8=-1.\]
Find the displacement over six seconds.
\[s(0)=8,\]
\[s(6)=36-36+8=8.\]
Therefore:
\[\text{Displacement}=8-8=0\text{ m}.\]
The object returns to its initial position.
Find the total distance traveled.
It travels from position
\(8\)
to position
\(-1\)
then back to
\(8\)
\[\text{Distance}=|{-1}-8|+|8-(-1)|.\]
\[\text{Distance}=9+9=18\text{ m}.\]
Displacement and distance are different quantities.
11. Mixed graph and derivative questions
Reading a derivative graph
A derivative graph gives slopes of the original function. It does not directly give the original function’s heights. Suppose a derivative graph shows:
\[f'(x)>0
\quad\text{for}\quad
x<-1,\]
\[f'(x)<0
\quad\text{for}\quad
-1<x<2,\]
\[f'(x)>0
\quad\text{for}\quad
x>2.\]
Assume
\(f\)
is continuous and differentiable at the two separating inputs, with:
\[f'(-1)=f'(2)=0.\]
Then:
- \(f\) increases before \(-1\).
- \(f\) decreases between \(-1\) and \(2\).
- \(f\) increases after \(2\).
- There is a local maximum at \(x=-1\).
- There is a local minimum at \(x=2\).
However, the maximum and minimum values cannot be calculated from these derivative signs alone. We need the original function or additional information about its values.
Reading the original graph
From a graph of
\(f\)
- Rising sections suggest positive derivative values.
- Falling sections suggest negative derivative values.
- Smooth horizontal tangents give zero derivative values.
- Sharp corners may indicate that the derivative does not exist.
A graph crossing the
\(x\)
-axis means:
\[f(x)=0,\]
not necessarily:
\[f'(x)=0.\]
12. Common mistakes
Mistake 1: Assuming every stationary point is an extremum
A zero derivative identifies a candidate. Check the derivative’s sign on both sides.
Mistake 2: Using \(f(x)\) instead of \(f'(x)\) for monotonicity
The sign of
\(f(x)\)
tells us whether the graph is above or below the axis.
The sign of
\(f'(x)\)
tells us its tangent direction.
Mistake 3: Forgetting endpoints
An absolute maximum or minimum on a closed interval may occur at an endpoint.
Mistake 4: Ignoring the domain
An undefined input is not a point on the graph. It may divide the domain, but it cannot itself be a maximum, minimum, or critical point.
Mistake 5: Giving an input when a value is requested
If a minimum occurs at:
\[(3,-4),\]
the minimum value is
\(-4\)
not
\(3\)
Mistake 6: Confusing rest with reversal
An object can have zero velocity momentarily and continue in the same direction afterward. A direction change requires a change in the sign of velocity.
Mistake 7: Confusing displacement with distance
Displacement is net position change. Distance adds the lengths of all parts of the journey.
13. Lesson Review
You should now be able to:
- Use derivative signs to identify increasing and decreasing intervals.
- Distinguish stationary points from other critical points.
- Apply the first-derivative sign test.
- Calculate local maximum and minimum values.
- Check endpoints when finding absolute extrema on closed intervals.
- Build and solve simple optimization models.
- Interpret velocity as the derivative of position.
- Distinguish position, displacement, distance, velocity, and speed.
- Read information about a function from its derivative graph.
The central relationships are:
\[f'(x)>0
\quad\Longrightarrow\quad
\text{increasing behavior on an interval},\]
\[f'(x)<0
\quad\Longrightarrow\quad
\text{decreasing behavior on an interval},\]
and:
\[v(t)=s'(t).\]