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Introduction
1. Limits
2. Derivative basics
3. Advanced differentiation
4. Contextual uses
5. Analytical uses
5.1 Important theorems
5.2 1st derivative test
5.3 2nd derivative test
5.4 Graphs & curve sketching
5.5 Optimization
6. Integration
7. Differential equations
8. Applications of integrals
9. Testing details tag
Wrapping up
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5.4 Graphs & curve sketching
Achievable AP Calculus AB
5. Analytical uses
Our AP Calculus AB course is currently in development and is a work-in-progress.

Graphs & curve sketching

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What you’ll learn:

  • How to interpret and connect the graphs of f,f′, and f′′ to analyze function behavior
  • How to sketch curves by combining domain, intercepts, asymptotes, extrema, and concavity

To analyze a function, you need to connect what you see on the graph of f to what its derivatives say about slope and concavity. On the AP exam, many multiple-choice questions on this topic focus more on reasoning from these connections than on computing exact values.

Connecting graphs of f,f′, and f′′

Recall from the previous sections:

The 1st derivative f′(x) tells you about the slope of the tangent line to f(x) at any x.

  • If f′(x)>0, f(x) is increasing.
  • If f′(x)<0, f(x) is decreasing.
  • If f′(a)=0 or f′(a) is undefined, then x=a is a critical point.
  • f(a) is an extrema if f′(x) changes sign around the critical point x=a.

The 2nd derivative f′′(x) tells you about the concavity of f(x).

  • When f′′(x)>0, f(x) is concave up and f′(x) is increasing.
  • When f′′(x)<0, f(x) is concave down and f′(x) is decreasing.
  • x=a is a point of inflection if f′′(x) changes sign around the point.

Examples

Shown below is the graph of f′(x), the derivative of f. The tangent lines at x=−1 and x=2 are horizontal.

Figure 5.4.1 Graph of f'
Figure 5.4.1 Graph of f'

Determine which of the 4 graphs A through D could be the graph of f(x).

Choices for f(x)
Choices for f(x)

First, locate the critical points of f(x). These occur where f′(x)=0 or where f′(x) is undefined.

Since the graph of f′(x) crosses the x-axis at x=−2,1, and 3, those are critical points of f(x). Now classify each one using the 1st derivative test (look for sign changes in f′):

  • At x=−2: f′(x) changes from negative to positive → relative min at x=−2.
  • At x=1: f′(x) changes from positive to negative → relative max at x=1.
  • At x=3: f′(x) changes from negative to positive → relative min at x=3.

You can also justify the same classifications using the 2nd derivative test by reading whether f′(x) is increasing or decreasing at each critical point:

  • At x=−2: f′(x) is increasing, so f′′(−2)>0→ relative min
  • At x=1: f′(x) is decreasing, so f′′(1)<0→ relative max
  • At x=3: f′(x) is increasing, so f′′(3)>0→ relative min

Only graphs B and C show extrema in this order, so A and D can be eliminated.

Now compare B and C more carefully. Graph B is made of straight-line segments with sharp corners. That would make f′(x) constant on each segment and undefined at the corner points x=−2,1, and 3. But the given graph of f′(x) shows f′(−2)=0, f′(1)=0, and f′(3)=0 (not undefined).

So graph B can be eliminated, and graph C is the only option.

Shown is the graph of a twice-differentiable function f(x) with. its extrema at x=−3 and x=1 (for clarity, the inflection points occur at x=−1.4 and x=2.75).

At which point is f′(x)<f(x)<f′′(x)?

a) x=−3
b) x=−1
c) x=1
d) x=3

Graph of f
Graph of f

Solution

(spoiler)

Answer: x=2

We need a point where f′(x) is the smallest value, f(x) is in the middle, and f′′(x) is the largest value.

At x=−3, f(−3)<0 (the graph is below the x-axis), but x=−3 is a relative min, so f′(−3)=0. Since f(−3)<f′(−3), this answer choice can be eliminated.

At x=−1, f(−1)=0, but the function is increasing there, so f′(−1)>0. Since f(−1)<f′(−1), this answer choice can also be eliminated.

At x=1, f(1)>0 and x=1 is a relative max, so f′(1)=0. Also, the graph is concave down there, so f′′(1)<0. That gives f′′(1)<f′(1)<f(1), so choice x=1 can be eliminated.

At x=3, f(3)=0. The function is decreasing, so f′(3)<0. The graph is concave up, so f′′(3)>0. This matches f′(3)<f(3)<f′′(3), so this is the correct point.

Curve sketching

You may have sketched general shapes of polynomial, rational, exponential, and other functions in Precalculus by finding intercepts and end behavior. With calculus, you can sketch more accurate graphs because derivatives tell you where the function increases/decreases and where it changes concavity.

To sketch the shape of a function:

  1. Find domain, x and y-intercepts, and asymptotes (if any)
  2. Find critical points (1st derivative f′)
  3. Test intervals for increasing/decreasing behavior of f and/or classify extrema
  4. Find inflection points (2nd derivative f′′)
  5. Test intervals for concavity of f
  6. Piece it all together

Sketch the graph of

f(x)=x2−1x​

Solution

(spoiler)

Step 1: Asymptotes, intercepts, domain

  • Vertical asymptotes: x=1 and x=−1
  • Horizontal asymptote: y=0
  • x-intercepts (when y=0): (0,0)
  • y-intercept (when x=0): (0,0)
  • Domain: (−∞,−1)∪(−1,1)∪(1,∞)

Step 2: 1st derivative

Start with the quotient rule:

f′(x)=(x2−1)2(x2−1)(1)−(x)(2x)​

=(x2−1)2x2−1−2x2​

=(x2−1)2−x2−1​

Now identify where f′(x) is zero or undefined.

f′(x) is undefined when x=−1 and x=1. These are vertical asymptotes where f(x) is undefined, so they are not critical points of f.

f′(x)=0 when −x2−1=0, which has no real solutions. So there are no relative extrema.

Step 3: Sign of f′(x)

Choose a test point in each interval of the domain (split by the vertical asymptotes at x=−1 and x=1) and determine the sign of f′(x). Here’s the sign diagram:

Sign diagram for f'(x)
Sign diagram for f'(x)

Since the 1st derivative is negative, f(x) is decreasing on all three intervals of its domain. When you sketch, it often helps to mark where f is undefined (open circles at the asymptotes) and to draw arrows showing the direction of decrease.

Step 4: 2nd derivative

Differentiate f′(x) to find f′′(x):

f′′(x)=(x2−1)4(x2−1)2(−2x)−(−x2−1)(2(x2−1)(2x))​

=(x2−1)4−2x(x2−1)−4x(x2−1)(−x2−1)​

Simplify by factoring −2x(x2−1) from the numerator:

f′′(x)=(x2−1)4−2x(x2−1)[1+2(−x2−1)]​

=(x2−1)4−2x(x2−1)(−2x2−1)​

Potential inflection points occur where f′′(x)=0 (and f is defined). Here, f′′(x)=0 when x=0 only.

Step 5: Determine concavity

Test the sign of f′′(x) on each interval. Even though x=−1 and x=1 cannot be inflection points (the function isn’t defined there), they still split the domain and should appear on the sign diagram.

Sign diagram for f''(x)
Sign diagram for f''(x)

As shown, it can be helpful to sketch a small “concave up” or “concave down” curve on each interval to keep the shape straight as you draw the final graph.

Step 6: Putting it all together

Combine what you know about intercepts, asymptotes, increasing/decreasing behavior, and concavity. For example, on (1,∞), f(x) is decreasing and concave up, so the right-hand branch must slope downward while bending upward.

f, f', and f''
f, f', and f''

Plot the asymptotes and intercepts, then sketch each branch to match the sign diagrams:

Final graph of f(x)
Final graph of f(x)

Connecting graphs of f, f′, and f′′

  • f′(x): slope of f(x); sign indicates increasing/decreasing; zeros/undefined = critical points
  • f′′(x): concavity of f(x); sign indicates concave up/down; sign changes = inflection points
  • Extrema at critical points where f′(x) changes sign; inflection points where f′′(x) changes sign

Analyzing f′(x) to determine f(x)

  • Critical points where f′(x)=0 or undefined
  • 1st derivative test: sign change in f′(x) classifies extrema (min/max)
  • 2nd derivative test: f′′(x)>0 at critical point = min, f′′(x)<0 = max

Comparing f(x), f′(x), and f′′(x) at points

  • Use graph features to estimate values and order of f(x), f′(x), f′′(x)
  • At extrema: f′(x)=0; at inflection: f′′(x)=0
  • Concavity and slope inform relative sizes

Curve sketching process

  • Step 1: Find domain, intercepts, asymptotes
  • Step 2: Find critical points using f′(x)
  • Step 3: Test intervals for increasing/decreasing (f′(x) sign)
  • Step 4: Find inflection points using f′′(x)
  • Step 5: Test intervals for concavity (f′′(x) sign)
  • Step 6: Combine all info to sketch graph

Example: f(x)=x2−1x​

  • Vertical asymptotes: x=1, x=−1; horizontal asymptote: y=0
  • Intercepts: (0,0); domain excludes x=±1
  • f′(x)=(x2−1)2−x2−1​: always negative, so f(x) always decreasing
  • No real critical points (no relative extrema)
  • f′′(x)=(x2−1)4−2x(x2−1)(−2x2−1)​; inflection point at x=0
  • Concavity changes at x=0; use sign diagrams for f′(x) and f′′(x) to guide sketch

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Graphs & curve sketching

What you’ll learn:

  • How to interpret and connect the graphs of f,f′, and f′′ to analyze function behavior
  • How to sketch curves by combining domain, intercepts, asymptotes, extrema, and concavity

To analyze a function, you need to connect what you see on the graph of f to what its derivatives say about slope and concavity. On the AP exam, many multiple-choice questions on this topic focus more on reasoning from these connections than on computing exact values.

Connecting graphs of f,f′, and f′′

Recall from the previous sections:

The 1st derivative f′(x) tells you about the slope of the tangent line to f(x) at any x.

  • If f′(x)>0, f(x) is increasing.
  • If f′(x)<0, f(x) is decreasing.
  • If f′(a)=0 or f′(a) is undefined, then x=a is a critical point.
  • f(a) is an extrema if f′(x) changes sign around the critical point x=a.

The 2nd derivative f′′(x) tells you about the concavity of f(x).

  • When f′′(x)>0, f(x) is concave up and f′(x) is increasing.
  • When f′′(x)<0, f(x) is concave down and f′(x) is decreasing.
  • x=a is a point of inflection if f′′(x) changes sign around the point.

Examples

Shown below is the graph of f′(x), the derivative of f. The tangent lines at x=−1 and x=2 are horizontal.

Determine which of the 4 graphs A through D could be the graph of f(x).

First, locate the critical points of f(x). These occur where f′(x)=0 or where f′(x) is undefined.

Since the graph of f′(x) crosses the x-axis at x=−2,1, and 3, those are critical points of f(x). Now classify each one using the 1st derivative test (look for sign changes in f′):

  • At x=−2: f′(x) changes from negative to positive → relative min at x=−2.
  • At x=1: f′(x) changes from positive to negative → relative max at x=1.
  • At x=3: f′(x) changes from negative to positive → relative min at x=3.

You can also justify the same classifications using the 2nd derivative test by reading whether f′(x) is increasing or decreasing at each critical point:

  • At x=−2: f′(x) is increasing, so f′′(−2)>0→ relative min
  • At x=1: f′(x) is decreasing, so f′′(1)<0→ relative max
  • At x=3: f′(x) is increasing, so f′′(3)>0→ relative min

Only graphs B and C show extrema in this order, so A and D can be eliminated.

Now compare B and C more carefully. Graph B is made of straight-line segments with sharp corners. That would make f′(x) constant on each segment and undefined at the corner points x=−2,1, and 3. But the given graph of f′(x) shows f′(−2)=0, f′(1)=0, and f′(3)=0 (not undefined).

So graph B can be eliminated, and graph C is the only option.

Shown is the graph of a twice-differentiable function f(x) with. its extrema at x=−3 and x=1 (for clarity, the inflection points occur at x=−1.4 and x=2.75).

At which point is f′(x)<f(x)<f′′(x)?

a) x=−3
b) x=−1
c) x=1
d) x=3

Solution

(spoiler)

Answer: x=2

We need a point where f′(x) is the smallest value, f(x) is in the middle, and f′′(x) is the largest value.

At x=−3, f(−3)<0 (the graph is below the x-axis), but x=−3 is a relative min, so f′(−3)=0. Since f(−3)<f′(−3), this answer choice can be eliminated.

At x=−1, f(−1)=0, but the function is increasing there, so f′(−1)>0. Since f(−1)<f′(−1), this answer choice can also be eliminated.

At x=1, f(1)>0 and x=1 is a relative max, so f′(1)=0. Also, the graph is concave down there, so f′′(1)<0. That gives f′′(1)<f′(1)<f(1), so choice x=1 can be eliminated.

At x=3, f(3)=0. The function is decreasing, so f′(3)<0. The graph is concave up, so f′′(3)>0. This matches f′(3)<f(3)<f′′(3), so this is the correct point.

Curve sketching

You may have sketched general shapes of polynomial, rational, exponential, and other functions in Precalculus by finding intercepts and end behavior. With calculus, you can sketch more accurate graphs because derivatives tell you where the function increases/decreases and where it changes concavity.

To sketch the shape of a function:

  1. Find domain, x and y-intercepts, and asymptotes (if any)
  2. Find critical points (1st derivative f′)
  3. Test intervals for increasing/decreasing behavior of f and/or classify extrema
  4. Find inflection points (2nd derivative f′′)
  5. Test intervals for concavity of f
  6. Piece it all together

Sketch the graph of

f(x)=x2−1x​

Solution

(spoiler)

Step 1: Asymptotes, intercepts, domain

  • Vertical asymptotes: x=1 and x=−1
  • Horizontal asymptote: y=0
  • x-intercepts (when y=0): (0,0)
  • y-intercept (when x=0): (0,0)
  • Domain: (−∞,−1)∪(−1,1)∪(1,∞)

Step 2: 1st derivative

Start with the quotient rule:

f′(x)=(x2−1)2(x2−1)(1)−(x)(2x)​

=(x2−1)2x2−1−2x2​

=(x2−1)2−x2−1​

Now identify where f′(x) is zero or undefined.

f′(x) is undefined when x=−1 and x=1. These are vertical asymptotes where f(x) is undefined, so they are not critical points of f.

f′(x)=0 when −x2−1=0, which has no real solutions. So there are no relative extrema.

Step 3: Sign of f′(x)

Choose a test point in each interval of the domain (split by the vertical asymptotes at x=−1 and x=1) and determine the sign of f′(x). Here’s the sign diagram:

Since the 1st derivative is negative, f(x) is decreasing on all three intervals of its domain. When you sketch, it often helps to mark where f is undefined (open circles at the asymptotes) and to draw arrows showing the direction of decrease.

Step 4: 2nd derivative

Differentiate f′(x) to find f′′(x):

f′′(x)=(x2−1)4(x2−1)2(−2x)−(−x2−1)(2(x2−1)(2x))​

=(x2−1)4−2x(x2−1)−4x(x2−1)(−x2−1)​

Simplify by factoring −2x(x2−1) from the numerator:

f′′(x)=(x2−1)4−2x(x2−1)[1+2(−x2−1)]​

=(x2−1)4−2x(x2−1)(−2x2−1)​

Potential inflection points occur where f′′(x)=0 (and f is defined). Here, f′′(x)=0 when x=0 only.

Step 5: Determine concavity

Test the sign of f′′(x) on each interval. Even though x=−1 and x=1 cannot be inflection points (the function isn’t defined there), they still split the domain and should appear on the sign diagram.

As shown, it can be helpful to sketch a small “concave up” or “concave down” curve on each interval to keep the shape straight as you draw the final graph.

Step 6: Putting it all together

Combine what you know about intercepts, asymptotes, increasing/decreasing behavior, and concavity. For example, on (1,∞), f(x) is decreasing and concave up, so the right-hand branch must slope downward while bending upward.

Plot the asymptotes and intercepts, then sketch each branch to match the sign diagrams:

Key points

Connecting graphs of f, f′, and f′′

  • f′(x): slope of f(x); sign indicates increasing/decreasing; zeros/undefined = critical points
  • f′′(x): concavity of f(x); sign indicates concave up/down; sign changes = inflection points
  • Extrema at critical points where f′(x) changes sign; inflection points where f′′(x) changes sign

Analyzing f′(x) to determine f(x)

  • Critical points where f′(x)=0 or undefined
  • 1st derivative test: sign change in f′(x) classifies extrema (min/max)
  • 2nd derivative test: f′′(x)>0 at critical point = min, f′′(x)<0 = max

Comparing f(x), f′(x), and f′′(x) at points

  • Use graph features to estimate values and order of f(x), f′(x), f′′(x)
  • At extrema: f′(x)=0; at inflection: f′′(x)=0
  • Concavity and slope inform relative sizes

Curve sketching process

  • Step 1: Find domain, intercepts, asymptotes
  • Step 2: Find critical points using f′(x)
  • Step 3: Test intervals for increasing/decreasing (f′(x) sign)
  • Step 4: Find inflection points using f′′(x)
  • Step 5: Test intervals for concavity (f′′(x) sign)
  • Step 6: Combine all info to sketch graph

Example: f(x)=x2−1x​

  • Vertical asymptotes: x=1, x=−1; horizontal asymptote: y=0
  • Intercepts: (0,0); domain excludes x=±1
  • f′(x)=(x2−1)2−x2−1​: always negative, so f(x) always decreasing
  • No real critical points (no relative extrema)
  • f′′(x)=(x2−1)4−2x(x2−1)(−2x2−1)​; inflection point at x=0
  • Concavity changes at x=0; use sign diagrams for f′(x) and f′′(x) to guide sketch