Applications of Differentiation: Curve Sketching

We say that a function $f(x)$ is increasing in an interval if the graph continuously rises as $x$ goes from left to right through the interval. That is, whenever $x_1$ and $x_2$ are in the interval with $x_1 < x_2$, we have $f(x_1) < f(x_2)$. We say that $f(x)$ is increasing at $x = c$ provided that $f(x)$ is increasing in some open interval on the $x$-axis that contains the point $c$.

We say that a function $f(x)$ is decreasing in an interval provided that the graph continuously falls as $x$ goes from left to right through the interval. That is, whenever $x_1$ and $x_2$ are in the interval with $x_1 < x_2$, we have $f(x_1) > f(x_2)$. We say that $f(x)$ is decreasing at $x = c$ provided that $f(x)$ is decreasing in some open interval that contains the point $c$.

A relative maximum point is a point on the graph whose height is at least as large as the heights of nearby points. A relative minimum point is a point on the graph whose height is at most as large as the heights of nearby points.

The maximum value (or absolute maximum value) of a function is the largest value that the function assumes on its domain. The minimum value (or absolute minimum value) of a function is the smallest value that the function assumes on its domain.

When a drug is injected intramuscularly (into a muscle), the concentration of the drug in the veins has the time-concentration curve shown below. We can describe this graph using the terms introduced previously.

Initially (when $t = 0$), there is no drug in the veins. When the drug is injected into the muscle, it begins to diffuse into the bloodstream. The drug concentration in the veins increases until it reaches its absolute maximum value at $t = 2$ hours.

After this time, the concentration begins to decrease as the body’s metabolic processes remove the drug from the blood. Eventually, the drug concentration decreases to a level so small that, for all practical purposes, it is zero. The peak at $t=2$ represents the relative and absolute maximum of this model.

A British manufacturer models the profit $P(x)$ from producing and selling $x$ thousand electric kettles. The profit function, shown below, accounts for both the revenue generated from sales and the costs associated with production.

At low production levels, the profit is negative (a loss) because the initial fixed overhead costs outweigh the revenue. As production increases, $P(x)$ increases, eventually turning positive and crossing the break-even point.

The profit continues to rise until production reaches $x = 4$ thousand units. At this point, the profit achieves an absolute maximum. If production is pushed beyond this level, inefficiencies and escalating marginal costs cause the profit to decrease. The relative maximum at $x=4$ represents the best production target for the manufacturer.

The daily number of hours of daylight in London increases from about $7.8$ hours on 21 December to about $12.2$ hours on 21 March, and then to about $16.6$ hours on 21 June. From 21 December to 21 March, the daily increase is greater than the previous daily increase, and from 21 March to 21 June, the daily increase is less than the previous daily increase.

If we draw a possible graph of the number of hours of daylight as a function of time, the first part of the graph is increasing at an increasing rate. The second part of the graph is increasing at a decreasing rate.

We say that a function $f(x)$ is concave up at $x = a$ if there is an open interval on the $x$-axis containing $a$ throughout which the graph of $f(x)$ lies above its tangent line. Equivalently, $f(x)$ is concave up at $x = a$ if the slope of the graph increases as we move from left to right through $(a, f(a))$.

We say that a function $f(x)$ is concave down at $x = a$ if there is an open interval on the $x$-axis containing $a$ throughout which the graph of $f(x)$ lies below its tangent line. Equivalently, $f(x)$ is concave down at $x = a$ if the slope of the graph decreases as we move from left to right through $(a, f(a))$.

An inflection point is a point on the graph of a function at which the function is continuous and at which the graph changes from concave up to concave down or vice versa.

Use the terms defined earlier to describe the graph shown below.

1. For $x < 3$, $f(x)$ is increasing and concave down.
2. Relative maximum point at $x = 3$.
3. For $3 < x < 4$, $f(x)$ is decreasing and concave down.
4. Inflection point at $x = 4$.
5. For $4 < x < 5$, $f(x)$ is decreasing and concave up.
6. Relative minimum point at $x = 5$.
7. For $x > 5$, $f(x)$ is increasing and concave up.

If $f'(x) > 0$ throughout an interval, then $f$ is increasing on that interval.

If $f'(x) < 0$ throughout an interval, then $f$ is decreasing on that interval.

Sketch the graph of a function $f(x)$ satisfying all of the following:

1. $f(3) = 4$

2. $f'(x) > 0$ for $x < 3$, $\quad f'(3) = 0$, $\quad f'(x) < 0$ for $x > 3$

We begin with the single concrete point $(3, 4)$ and draw a horizontal tangent there since $f'(3) = 0$. By the first derivative rule, $f$ is increasing for every $x < 3$ and decreasing for every $x > 3$. The graph rises to $(3, 4)$ and falls away from it on both sides, making $(3, 4)$ a relative maximum.

If $f''(x) > 0$ throughout an interval, then $f$ is concave up on that interval.

If $f''(x) < 0$ throughout an interval, then $f$ is concave down on that interval.

Sketch the graph of a function $f$ with the following properties:

1. The points $(2, 3)$, $(4, 5)$, and $(6, 7)$ lie on the graph.

2. $f'(2) = 0$ and $f'(6) = 0$.

3. $f''(x) > 0$ for $x < 4$, $\quad f''(4) = 0$, $\quad f''(x) < 0$ for $x > 4$.

The second derivative rule, combined with the zero-derivative conditions, determines the character of each key point. At $(2, 3)$: $f'(2) = 0$ and $f''(2) > 0$ (since $2 < 4$), so a horizontal tangent on a concave-up curve — this is a relative minimum. At $(6, 7)$: $f'(6) = 0$ and $f''(6) < 0$ (since $6 > 4$), so a horizontal tangent on a concave-down curve — a relative maximum. At $x = 4$: concavity changes from up to down, so $(4, 5)$ is an inflection point.

Between the two extrema the function increases through a concave-up bowl (for $x < 4$) into a concave-down arch (for $x > 4$), crossing through the inflection point at $(4, 5)$ where the transition occurs.

The function $f(x) = 8x - x^2$ is graphed below together with the slope at several points.

The slopes are decreasing as we move from left to right: steeply positive at $x = 1$, zero at $x = 4$, and steeply negative at $x = 7$. So $f'(x)$ is a decreasing function. Computing $f'(x) = 8 - 2x$, whose graph is shown below.

The $y$-values of $f'$ match the slopes on the parabola exactly: positive for $x < 4$, zero at $x = 4$, negative for $x > 4$. The parabola is the shape of a typical revenue curve; the graph of $f'$ is the corresponding marginal revenue curve, whose zero at $x = 4$ identifies the revenue-maximising output.

The function $f(x) = \tfrac{1}{3}x^3 - 4x^2 + 18x + 10$ models a total cost curve whose slope — the marginal cost — first decreases and then increases. Several slopes are marked on the graph of $f$ below.

The derivative is $f'(x) = x^2 - 8x + 18$, an upward-opening parabola graphed below.

The $y$-values on the graph of $f'$ decrease and then increase, reaching their minimum at $x = 4$. Since $(f')'(x) = f''(x) = 2x - 8$, we have $f''(4) = 0$, and $f''$ changes sign there from negative to positive. By the second derivative rule, $x = 4$ is the inflection point of $f$ — where the cost curve transitions from concave down to concave up. The inflection point of the cost curve coincides exactly with the input at which the marginal cost curve is minimised.

The graph of $y = f'(x)$ is shown below.

1. What is the slope of $f$ at $x = 1$?

Since $f'(1) = 2$, the graph of $f$ has slope $2$ at $x = 1$.

2. Describe how $f'(x)$ changes on $1 \leq x \leq 2$.

The values of $f'(x)$ are positive and decreasing as $x$ increases from $1$ to $2$.

3. Describe the shape of $f$ on $1 \leq x \leq 2$.

Since $f'(x) > 0$ throughout this interval, $f$ is increasing there. Since $f'$ is decreasing on this interval, the slope of $f$ is decreasing, which by the Concavity definition means $f$ is concave down on $1 \leq x \leq 2$.

4. For what values of $x$ does $f$ have a horizontal tangent?

Horizontal tangents occur where $f'(x) = 0$. From the graph, these are $x=-1$ and $x=3$.

5. Explain why $f$ has a relative maximum at $x = 3$.

Just to the left of $x = 3$, $f'(x) > 0$, so $f$ is increasing; just to the right, $f'(x) < 0$, so $f$ is decreasing. By the Relative Maximum definition, $f$ transitions from increasing to decreasing at $x = 3$ and therefore has a relative maximum there.

A graph of $f$ consistent with these conditions is shown below.

A number $a$ in the domain of $f$ is a critical number of $f$ if $f'(a) = 0$ or if $f'(a)$ does not exist. The point $(a, f(a))$ on the graph is called a critical point.

The first derivative rule ensures that every local extreme point occurs at a critical number. The converse fails: a critical number is a candidate, and whether it is an extremum depends on the sign change of $f'$ there.

Suppose $a$ is a critical number of $f$, and suppose $f'$ has a definite sign on each side of $a$ near $a$.

1. If $f'$ changes from positive to negative at $a$, then $f$ has a local maximum at $a$.

2. If $f'$ changes from negative to positive at $a$, then $f$ has a local minimum at $a$.

3. If $f'$ does not change sign at $a$, then $f$ has no local extremum at $a$.

For $f(x) = |x|$, the derivative is $f'(x) = -1$ for $x < 0$ and $f'(x) = 1$ for $x > 0$, but $f'(0)$ does not exist because the left and right slopes disagree. The input $0$ is still a critical number because $f(0)$ is defined.

The sign of $f'$ changes from negative on the left of $0$ to positive on the right of $0$, so the first derivative test gives a local minimum at $(0,0)$. This is why the definition of critical number includes both $f'(a)=0$ and the case where $f'(a)$ does not exist.

Find the local maximum and minimum points of $f(x) = \tfrac{1}{3}x^3 - 2x^2 + 3x + 1$.

Step 1: find critical numbers. Differentiating,

Setting $f'(x) = 0$ gives $x = 1$ and $x = 3$.

Step 2: evaluate $f$ at the critical numbers.

The critical points are $\bigl(1, \tfrac{7}{3}\bigr)$ and $(3, 1)$.

Step 3: sign chart. The critical numbers divide the real line into three intervals. Sampling each factor at a representative input from each interval:

The sign of $f'$ changes from $+$ to $-$ at $x = 1$ (local maximum) and from $-$ to $+$ at $x = 3$ (local minimum).

Find the local maximum and minimum points of $f(x) = (3x - 1)^3$.

Differentiating by the General power rule from Recitation 2,

Setting $f'(x) = 0$ gives $3x - 1 = 0$, so the only critical number is $x = \tfrac{1}{3}$, with $f\bigl(\tfrac{1}{3}\bigr) = 0$.

Since $(3x - 1)^2$ is a perfect square, it is non-negative for all $x$ and equals zero only at $x = \tfrac{1}{3}$. So $f'(x) > 0$ on both sides of $x = \tfrac{1}{3}$: the sign does not change. By case (c) of the first derivative test, there is no local extremum at $\bigl(\tfrac{1}{3}, 0\bigr)$. The function is increasing on its entire domain; the slope only pauses at zero before continuing upward.

After a patient takes a dose, the amount of drug in the bloodstream (in mg) is modelled for the first six hours by

The Real-World Applications section described the shape of this type of curve — rising to a peak, then falling as the body clears the drug. The first derivative test now locates that peak exactly.

Differentiating,

Setting $A'(t) = 0$ gives $t = 0$ and $t = 4$. The value $t=0$ is the left end of the time interval, while $t=4$ is the turning point inside the interval. For $0<t<4$, the factor $-6t$ is negative and the factor $(t-4)$ is negative, so $A'(t)>0$ and $A$ is increasing. For $4<t<6$, the factor $-6t$ is negative and $(t-4)$ is positive, so $A'(t)<0$ and $A$ is decreasing. The sign changes from positive to negative at $t = 4$: a local maximum.

The peak concentration is

Let $a$ be a critical number of $f$ with $f'(a) = 0$.

1. If $f''(a) < 0$, then $f$ has a local maximum at $a$.

2. If $f''(a) > 0$, then $f$ has a local minimum at $a$.

3. If $f''(a) = 0$, the test is inconclusive: $f$ may have a local maximum, a local minimum, or no extremum at $a$.

Locate the extreme point of $f(x) = \tfrac{1}{4}x^2 - x + 2$.

Setting $f'(x) = 0$ gives $x = 2$, the only critical number, with $f(2) = 1 - 2 + 2 = 1$. Since $f''(x) = \tfrac{1}{2} > 0$ everywhere, the second derivative test gives a local minimum at $(2, 1)$.

Because $f''$ is positive everywhere, the graph is concave up throughout; no other critical numbers exist, so this is also the absolute minimum.

Locate all extreme points of $f(x) = x^3 - 3x^2 + 5$.

The critical numbers are $x = 0$ and $x = 2$. Evaluating:

Applying the second derivative test:

Find the inflection point of $f(x) = x^3 - 3x^2 + 5$, and explain why the graph between the two extreme points has exactly one shape transition.

From the previous example, $f''(x) = 6x - 6 = 6(x - 1)$. Setting $f''(x) = 0$ gives $x = 1$, with $f(1) = 1 - 3 + 5 = 3$.

• For $x < 1$: $f''(x) < 0$, so $f$ is concave down.
• For $x > 1$: $f''(x) > 0$, so $f$ is concave up.

The concavity reverses at $x = 1$, confirming an inflection point at $(1, 3)$.

The second derivative test found a local maximum at $(0, 5)$ (concave down) and a local minimum at $(2, 1)$ (concave up). Since $f''$ changes sign exactly once — at $x = 1$ between these two points — the concavity transitions through $(1, 3)$ and nowhere else. This rules out the “wiggly” alternative sketch that would require additional concavity changes.

A pop-up market stall runs over six operating days. As the number of promotional hours $x$ invested each day grows, the daily revenue (in hundreds of pounds) follows the model

The initial dip reflects setup costs absorbed in early promotion. The three-step procedure locates the turnaround and the peak.

Step 1: Critical points.

Critical numbers: $x = 1$ and $x = 5$. Values:

Step 2: Classify.

$f''(x) = -2x + 6$.

The stall operates at a loss at $x = 1$ (revenue is negative), crosses break-even as promotional effort grows, and peaks at $x = 5$.

Step 3: Inflection.

Setting $f''(x) = 0$ gives $x = 3$. Since $f''(2) = 2 > 0$ and $f''(4) = -2 < 0$, the concavity reverses at $x = 3$. The inflection point is $\bigl(3,\, 3\bigr)$.

At $x = 3$, the rate of revenue growth transitions from accelerating to decelerating — the stall is still gaining but the gains are shrinking. The full sketch connects all three plotted features.