B. Derivatives and Their Applications · Curve Sketching: Connecting f, f′ and f″Sketching the Graph of f′ from the Graph of f
B1.1About 70 minutes
Learning goals- I can sketch the graph of f′ from the graph of f, using zeros, signs and steepness.
- I can explain where f′ is undefined (corners and cusps).
- I can recognize points of inflection on the graph of f as turning points of f′.
- I can describe how vertical and horizontal translations of f affect f′.
Words to know
derivative function f′
The function that gives the slope of the tangent to y = f(x) at each x.
increasing on an interval
For any x₁ < x₂ in the interval, f(x₁) < f(x₂): the graph rises from left to right.
decreasing on an interval
For any x₁ < x₂ in the interval, f(x₁) > f(x₂): the graph falls from left to right.
point of inflection
A point on a continuous curve where the direction of bending changes (from ∩-shaped to ∪-shaped or the reverse). On a smooth graph it is where the slope is steepest compared with nearby points.
corner / cusp
A sharp point on a graph where the slopes from the left and right do not agree; the derivative does not exist there.
The slope tells a story
Imagine you are tobogganing down a long hill in January and someone films your height above the bottom every second. The graph of height is f. Now ask a different question at every instant: how steep is it right now? That steepness, recorded at every x, is the graph of the derivative f′.
**Remember from Derivatives and Differentiation Rules:** the derivative f′(x) is the slope of the tangent to y = f(x) at each x. Where the slope is positive the function increases, where it is negative the function decreases, and at a local maximum or minimum of a smooth curve the tangent is horizontal.
In this lesson you will turn that idea into a skill: given only the graph of f, with no equation, sketch the graph of f′. Along the way you will meet a new kind of point, the point of inflection.
The dictionary between f and f′
Read the graph of f from left to right and translate each feature:
- f is increasing → f′ is positive (graph of f′ above the x-axis).
- f is decreasing → f′ is negative (graph of f′ below the x-axis).
- f has a horizontal tangent (often a local maximum or minimum) → f′ is zero (graph of f′ meets the x-axis).
- f is steep → f′ is far from 0. f is nearly flat → f′ is close to 0.
- f is a straight line → f′ is constant (a horizontal line).
- f has a corner or cusp → f′ is undefined there.
For a polynomial, f′ has degree one less than f. A cubic has a quadratic derivative; a quartic has a cubic derivative.
Points of inflection
Look at f(x) = x³ − 3x near x = 0. To the left of 0 the graph bends like the top of a hill (it curves downward, like ∩). To the right it bends like a bowl (it curves upward, like ∪). The point where the bending changes is a point of inflection.
Here is how to spot one using slopes. As you move along the curve, the slope of f keeps falling until x = 0, then starts rising again. So at a point of inflection, the slope of f is at its steepest (most positive or most negative) compared with nearby points. On the graph of f′, that is a local maximum or local minimum of f′.
So: points of inflection of f ↔ turning points of f′. In the figure above, the inflection point of f at (0, 0) lines up with the lowest point of f′ at x = 0. You will define inflection points more carefully with the second derivative in the lesson Concavity and Points of Inflection.
A sketching routine
- Mark every x where f has a horizontal tangent: these are the zeros of f′.
- Mark every corner or cusp: f′ is undefined there (use open circles or a gap).
- Between those x-values, write + where f rises and − where f falls.
- Find where f is steepest: these are the high or low points of f′ (the inflection points of f).
- Estimate one or two actual slopes (rise over run with a ruler) to set the vertical scale.
- Join everything with a smooth curve.
Worked exampleExample 1: Sketching f′ for a cubic
The graph of f(x) = x³ − 3x is shown in the first figure. Without using the equation, sketch the graph of f′. Then check your sketch with the equation.
- Horizontal tangents at x = −1 (top of the hill) and x = 1 (bottom of the valley). So f′(−1) = 0 and f′(1) = 0.
- f rises for x < −1, falls for −1 < x < 1, rises for x > 1. So f′ is +, then −, then +.
- f falls most steeply at x = 0, where the slope looks like about −3 (from (−0.5, 1.4) to (0.5, −1.4) the rise over run is −2.8 ÷ 1). So f′ has its lowest point near (0, −3).
- Far to the left and right f becomes very steep, so f′ grows large and positive at both ends.
- The shape is an upward-opening parabola through (−1, 0), (0, −3) and (1, 0).
- Check: f′(x) = 3x² − 3 = 3(x − 1)(x + 1). Its zeros are ±1 and f′(0) = −3. ✓
f′ is an upward parabola with zeros at x = −1 and x = 1 and vertex (0, −3). The vertex of f′ marks the point of inflection (0, 0) of f.
A quartic f with two local minimums and one local maximum has a cubic derivative with three zeros, at x = −2, 0 and 2. The signs of f′ go −, +, −, + from left to right, just as f goes down, up, down, up.Worked exampleExample 2: Sketching f′ for a quartic
The graph of f(x) = ¼x⁴ − 2x² has local minimums at (−2, −4) and (2, −4) and a local maximum at (0, 0). Sketch f′ from the graph.
- Horizontal tangents at x = −2, 0 and 2, so f′ has three zeros there.
- Reading left to right: f falls, rises, falls, rises. So f′ is −, +, −, +.
- A quartic has a cubic derivative. A cubic that starts negative and ends positive has a positive leading coefficient, which matches.
- f is steepest between each pair of turning points, at about x = −1.2 and x = 1.2. Those are the turning points of f′, and the inflection points of f.
- Check with the power rule: f′(x) = x³ − 4x = x(x − 2)(x + 2). ✓
f′ is a cubic with zeros at −2, 0 and 2, negative for x < −2, positive between −2 and 0, negative between 0 and 2, positive for x > 2.
At a corner the slope jumps from −1 to 1, so the derivative is undefined at x = 1. The open circles show that f′ has no value there.Worked exampleExample 3: A graph with a corner
Sketch the derivative of f(x) = |x − 1|.
- For x < 1, the graph is the line y = −(x − 1) = −x + 1, so its slope is −1.
- For x > 1, the graph is the line y = x − 1, with slope 1.
- At x = 1 the slope jumps. Secants from the left have slope −1 and from the right slope 1, so the limit that defines f′(1) does not exist (you saw this in Limits and Rates of Change).
f′(x) = −1 for x < 1 and f′(x) = 1 for x > 1, with no value at x = 1: two horizontal rays with open circles at x = 1.
Translations and the derivative
What happens to f′ when you translate f?
- Vertical translation g(x) = f(x) + k: every point moves up by k, but every tangent keeps its slope. So g′(x) = f′(x). The derivative cannot see a vertical shift.
- Horizontal translation g(x) = f(x − h): every feature, including every slope, moves h units right. So g′(x) = f′(x − h): the graph of the derivative shifts the same way.
The first fact has a big consequence. Many different functions share the same derivative. If you know only f′, you know the shape of f but not its height. You will use this idea in the lesson Reading f from Information About f′ and f″.
Worked exampleExample 4: Translations (a curriculum investigation)
Let f(x) = x³ − 3x. Compare the derivatives of g(x) = x³ − 3x + 5 and h(x) = (x − 2)³ − 3(x − 2).
- g is f moved up 5. Differentiate: g′(x) = 3x² − 3, the same as f′(x).
- h is f moved right 2. By the chain rule, h′(x) = 3(x − 2)² − 3, which is f′(x − 2).
- So the zeros of h′ are at x − 2 = ±1, that is x = 1 and x = 3: the zeros of f′ moved right 2.
g′ = f′ (no change). h′ is f′ shifted 2 units right, with zeros at 1 and 3.
Just for funA constant walks into a calculus class. The derivative looks at it and says, "Sorry, you're nothing to me." (The derivative of any constant is 0, which is exactly why vertical translations never change f′.)
Hands-on activitySpaghetti tangents
You need: one sheet of graph paper (or lined paper with a hand-drawn grid), a pencil, a ruler and one piece of uncooked spaghetti.
- Draw a smooth wavy curve across the page with at least one hill and one valley. Draw the x-axis below it with marks every 1 cm.
- At each centimetre mark, lay the spaghetti so it just touches the curve without crossing it. This is your tangent line.
- Estimate its slope: pick two points on the spaghetti, count the rise and run in centimetres, and divide. Record each slope in a table (positive, negative or about 0).
- On a new set of axes directly below, plot (x, slope) for every mark and join the points smoothly. That is your hand-made f′.
- Check the dictionary: your f′ should cross zero below each hill-top and valley-bottom, and its own turning points should sit below the places where your curve is steepest.
What to look for: the zeros of your f′ line up with the turning points of your curve within about half a centimetre. If they do not, re-check the slopes near those points; tangents near a turning point should look almost horizontal.
Your turn
1A smooth function f is increasing for x < −3, decreasing for −3 < x < 2 and increasing for x > 2. Describe where f′ is positive, negative and zero.
f′ > 0 for x < −3 and x > 2; f′ < 0 for −3 < x < 2; f′(−3) = 0 and f′(2) = 0.
2f(x) = −2x + 5. Describe the graph of f′.
The slope of the line is −2 everywhere, so f′ is the horizontal line y = −2.
3The graph of f is a parabola that opens down with vertex (4, 7). Describe the graph of f′.
A quadratic has a linear derivative. f rises left of 4 and falls right of 4, so f′ is a line with negative slope that is positive for x < 4, crosses the x-axis at x = 4, and is negative for x > 4.
4g(x) = f(x) + 10. How does the graph of g′ compare with the graph of f′?
They are identical. Moving a graph up 10 units does not change any tangent slope.
5h(x) = f(x + 4) and f′(2) = 5. For which x does h′(x) = 5, as a result?
h′(x) = f′(x + 4), so h′(x) = 5 when x + 4 = 2, that is x = −2. (h is f shifted 4 left, so its slopes shift 4 left too.)
6Describe the derivative of f(x) = |x + 2|.
f′(x) = −1 for x < −2 and 1 for x > −2. f′(−2) is undefined because of the corner at (−2, 0).
7The graph of f′ is a downward parabola with its highest point at x = 1. What feature does f have at x = 1?
f′ has a local maximum there, so f is rising (or falling) most steeply at x = 1 compared with nearby points: x = 1 is a point of inflection of f.
More practice: warm-up
1On an interval where f is increasing, is f′ positive or negative?
2f has a horizontal tangent at x = a. What is f′(a)?
3f(x) = 7. What is the graph of f′?
f′(x) = 0, so the graph of f′ is the x-axis.
4f is a cubic polynomial. What degree is f′?
5f′(x) = 3 for every x. What kind of graph does f have?
A straight line with slope 3.
More practice: on your own
1f(x) = x² − 4x. Sketch f′ and connect it to the vertex of f.
f′(x) = 2x − 4: a line with slope 2 crossing the x-axis at x = 2. The vertex of f is (2, −4); the horizontal tangent there is the zero of f′. f′ < 0 for x < 2 (f falls) and f′ > 0 for x > 2 (f rises).
2A cubic f with positive leading coefficient has a local maximum at (1, 6) and a local minimum at (5, −2). Describe f′ fully.
f′ is a quadratic with zeros at x = 1 and x = 5. f rises, falls, rises, so f′ is +, −, +: an upward parabola. By symmetry of a parabola its vertex is at x = 3, the point of inflection of f, where f falls most steeply.
3f(x) = √x for x ≥ 0. Describe f′ in words.
f increases everywhere, so f′ > 0. The curve gets flatter as x grows, so f′ decreases toward 0. Near x = 0 the curve is nearly vertical, so f′ grows without bound as x → 0⁺ and f′(0) is undefined. (Indeed f′(x) = 1/(2√x).)
4A graph is made of two straight pieces: slope −1 for x < 2 and slope 2 for x > 2, meeting at (2, 3). Describe f′.
f′(x) = −1 for x < 2 and 2 for x > 2; undefined at x = 2 (a corner). The graph of f′ is two horizontal rays with open circles at x = 2.
5f(x) = x³ and g(x) = (x − 1)³ + 4. Compare g′ with f′.
f′(x) = 3x². g′(x) = 3(x − 1)², which is f′ shifted 1 unit right. The +4 has no effect on the derivative. Both derivatives touch zero once (at x = 0 and x = 1) without changing sign.
Challenge
1Use the limit definition to prove that g(x) = f(x) + k has g′(x) = f′(x).
g′(x) = lim (h→0) [g(x + h) − g(x)]/h = lim [f(x + h) + k − f(x) − k]/h = lim [f(x + h) − f(x)]/h = f′(x). The k cancels in the numerator.
2f is an even function (f(−x) = f(x)) and differentiable. Show that f′ is odd. Check with f(x) = x⁴ − 3x².
Differentiate both sides of f(−x) = f(x) with the chain rule: −f′(−x) = f′(x), so f′(−x) = −f′(x): f′ is odd. Check: f′(x) = 4x³ − 6x, and f′(−x) = −4x³ + 6x = −f′(x). ✓
3Two functions both have derivative 2x + 1 for every x. One passes through (0, 3), the other through (1, 0). Find both equations.
Any function with derivative 2x + 1 is x² + x + C (they differ only by a vertical shift). Through (0, 3): C = 3, so f(x) = x² + x + 3. Through (1, 0): 1 + 1 + C = 0, C = −2, so g(x) = x² + x − 2.
Quiz
1. On an interval where f is decreasing, where does the graph of f′ lie?
2. A smooth function f has a local maximum at x = 4. What is f′(4)?
3. If g(x) = f(x) − 5, then g′(x) equals:
4. Where f increases most steeply (compared with nearby points), the graph of f′ has:
5. What is the derivative of f(x) = |x| at x = 0?
6. f is a quartic (degree 4) polynomial. What is f′?
Big idea (B1.1): the graph of f′ records the slope of f at every point. Your teen should be able to look at a graph with no equation and say where f′ is positive, negative, zero or undefined, and where f′ has its own highs and lows (the inflection points of f).
How to help: point to a spot on a graph and ask, "Is it going up or down here? Steep or gentle?" Then ask what the derivative's graph should be doing at that x. The spaghetti activity is worth doing together; it makes tangent lines physical.
Common difficulties:
- Copying the height of f instead of its slope. A high point on f is a zero of f′, not a high point.
- Forgetting that f′ is undefined at corners.
- Thinking that a vertical shift changes f′. It does not; a horizontal shift moves f′ sideways.
- Mixing up turning points of f (zeros of f′) with turning points of f′ (inflection points of f).