AP Calculus AB/BC · Unit 5 · Topic 5.8
Sketching Graphs of Functions and Their Derivatives
Transfer zeros, signs, slopes, extrema, and corners between f and f′ graphs.
1. Topic Focus
Use derivatives to prove existence, classify extrema, analyze monotonicity and concavity, sketch graphs, and solve optimization problems.
This topic: Transfer zeros, signs, slopes, extrema, and corners between f and f′ graphs.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
Where f rises, f′ is positive; horizontal tangents of f appear as zeros of f′.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
1. The derivative graph is a slope map
At each input \(x\), the height \(f'(x)\) equals the slope of the tangent to \(f\). Sketching \(f'\) from \(f\) means translating visible tangent slopes into heights; sketching \(f\) from \(f'\) means translating derivative heights back into direction and steepness.
2. From a graph of f to a graph of f'
Move from left to right along \(f\). Record where tangents are horizontal, positive, negative, steep, nearly flat, or undefined. Plot those slope values at the same inputs on the \(f'\) axes, then connect them consistently with the smoothness and domain of the original graph.
3. Horizontal tangents become zeros
A zero of \(f'\) can correspond to a local maximum, a local minimum, or a stationary point with no extremum. The sign pattern around the zero decides which occurs.
4. The sign of f' gives direction
Where \(f'>0\), the graph of \(f\) rises as \(x\) increases. Where \(f'<0\), it falls. Where \(f'=0\) throughout an interval, \(f\) is constant there.
5. Crossings of the derivative axis classify extrema
- Positive to negative: \(f\) has a local maximum.
- Negative to positive: \(f\) has a local minimum.
- No sign change: the horizontal tangent is not a local extremum.
6. Derivative magnitude gives steepness
A large \(|f'(x)|\) means \(f\) is steep; a small \(|f'(x)|\) means it is nearly horizontal. The sign gives direction, while the distance of the derivative graph from its horizontal axis gives the size of the slope.
7. The direction of f' gives concavity of f
If \(f'\) is increasing, tangent slopes increase and \(f\) is concave up. If \(f'\) is decreasing, tangent slopes decrease and \(f\) is concave down. This conclusion depends on whether \(f'\) rises or falls, not whether \(f'\) is above or below its axis.
8. Extrema of f' can signal inflection of f
A local extremum of \(f'\) is a candidate for a concavity change of \(f\). Confirm that \(f'\) changes from increasing to decreasing or vice versa and that the corresponding point belongs to the graph of \(f\).
9. Undefined derivatives create visible features
Corners, cusps, vertical tangents, and discontinuities can make \(f'\) undefined. A corner has different finite one-sided slopes, a cusp has slopes with opposite infinite behavior, and a vertical tangent has slopes with the same infinite direction. Inspect the original graph and one-sided behavior instead of automatically drawing one derivative point.
10. Preserve domain breaks
If \(f\) is discontinuous or undefined at an input, \(f'\) is also undefined there. Do not connect a derivative sketch across a hole or vertical asymptote unless the derivative itself is defined by separate information.
11. Derivative data do not fix vertical position
Functions that differ by a constant have the same derivative. Therefore \(f'\) determines the shape of \(f\) only up to a vertical translation. One value such as \(f(a)=b\) anchors the sketch at a definite height.
12. A reliable sketching workflow
- Align the horizontal scales of both graphs.
- Mark domain breaks and derivative-undefined inputs.
- Mark zeros of \(f'\) and classify sign changes.
- Label increasing and decreasing intervals of \(f\).
- Use the size of \(f'\) to vary steepness.
- Use the increase or decrease of \(f'\) to shape concavity.
- Anchor \(f\) with a known value when one is supplied.
13. Common errors
- Copying the graph of \(f\) as the graph of \(f'\).
- Using the height of \(f\) instead of its slope.
- Assuming every zero of \(f'\) creates an extremum.
- Confusing \(f'>0\) with concave up.
- Ignoring derivative magnitude when sketching steepness.
- Claiming a unique vertical position without an initial value.
6. Detailed Worked Example and Error Check
Example 1: Reconstructing from an algebraic derivative. Suppose \(f'(x)=(x+2)(x-1)\). Then \(f\) increases on \((-\infty,-2)\), decreases on \((-2,1)\), and increases on \((1,\infty)\). Thus \(f\) has a local maximum at \(x=-2\) and a local minimum at \(x=1\). Since \(f''(x)=2x+1\), the graph is concave down for \(x<-1/2\) and concave up for \(x>-1/2\).
Example 2: Sketching a derivative from a cubic. For \(f(x)=x^3-3x\),
The derivative is an upward-opening parabola with zeros at \(-1\) and \(1\). It is positive outside those inputs and negative between them, matching a local maximum of \(f\) at \(-1\) and a local minimum at \(1\). Its minimum at \(x=0\) matches the inflection input of \(f\).
Example 3: A corner. For \(f(x)=|x|\), the slope is \(-1\) for \(x<0\) and \(1\) for \(x>0\). The derivative graph consists of two horizontal pieces and has no value at zero. The sign change still shows that \(f\) has a local minimum at the corner.
Example 4: A vertical tangent without an extremum. For \(f(x)=x^{1/3}\), \(f'(x)=1/(3x^{2/3})\) is positive on both sides of zero and undefined at zero. The original graph keeps increasing through a vertical tangent, so zero is not a local extremum.
Example 5: A derivative that touches its axis. If \(f'(x)=x^2\), then \(f'>0\) on both sides of zero. Any antiderivative has the form \(f(x)=x^3/3+C\), so \(x=0\) is a stationary inflection input rather than an extremum.
Example 6: Trigonometric correspondence. For \(f(x)=\sin x\), the derivative is \(f'(x)=\cos x\). The maxima and minima of sine occur where cosine crosses zero, and the inflection inputs of sine occur where cosine has local extrema. The repeating derivative pattern preserves the period \(2\pi\).
Example 7: Reading a smooth derivative sketch. Suppose a smooth graph of \(f'\) passes through \((-2,-3),(-1,0),(0,2),(1,0),(2,-2)\), increases from \(x=-2\) to \(x=0\), and decreases from \(x=0\) to \(x=2\). Then \(f\) has a local minimum at \(x=-1\), a local maximum at \(x=1\), is concave up on \((-2,0)\), and is concave down on \((0,2)\). The graph of \(f\) has an inflection input at zero if it is continuous there.
7. AP Reasoning Routine
State theorem hypotheses, make sign charts on domain intervals, include endpoints when required, and connect derivative signs to function behavior.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Describe the requested sketch or behavior with enough information to reproduce it.
(a) If \(f'(x)=x(x-4)\), give the increasing and decreasing intervals and local extrema of \(f\).
(b) If \(f'(x)=(x+1)^2(x-2)\), determine the direction, local extrema, and concavity intervals of \(f\).
(c) For \(f(x)=x^4-4x^2\), describe the important features of the graph of \(f'\).
(d) Sketch the derivative behavior of \(f(x)=|x-2|\).
(e) Use \(f(x)=x^{2/3}\) to explain how a cusp appears in the derivative and classify the cusp.
(f) Given \(f'(x)=\cos x\) on \([0,2\pi]\), describe the shape and extrema of \(f\) up to vertical translation.
(g) A derivative is positive on \((-\infty,-2)\), negative on \((-2,1)\), and positive on \((1,\infty)\). Describe the local extrema of \(f\).
(h) A graph of \(f'\) increases on \((-3,0)\), decreases on \((0,4)\), and is continuous. State the concavity of \(f\) and the possible inflection input.
(i) If \(f'(x)=2x\), describe all possible functions \(f\), then identify the unique one satisfying \(f(0)=3\).
(j) Explain how the derivative sketch differs near a horizontal tangent, a corner, and a vertical tangent of \(f\).
Check the solution
In part (a), \(f'>0\) on \((-\infty,0)\) and \((4,\infty)\), and \(f'<0\) on \((0,4)\). Thus \(f\) has a local maximum at \(x=0\) and a local minimum at \(x=4\). In part (b), \(f'\) is negative for \(x<2\), except that it equals zero at \(-1\), and positive for \(x>2\). Thus \(-1\) is not an extremum input and \(x=2\) is a local minimum input. Since \(f'(x)=x^3-3x-2\), \(f''(x)=3(x-1)(x+1)\); \(f\) is concave up on \((-\infty,-1)\) and \((1,\infty)\), and concave down on \((-1,1)\). In part (c), \(f'(x)=4x(x^2-2)\) is a cubic crossing its axis at \(-\sqrt2,0,\sqrt2\). Its signs are negative, positive, negative, positive across those zeros, matching minima of \(f\) at \(\pm\sqrt2\) and a maximum at zero. In part (d), \(f'=-1\) for \(x<2\), \(f'=1\) for \(x>2\), and \(f'(2)\) is undefined. In part (e), \(f'(x)=2/(3x^{1/3})\) tends to negative infinity from the left and positive infinity from the right. The sign changes from negative to positive, so the cusp at \((0,0)\) is a local minimum. In part (f), an antiderivative is \(f(x)=\sin x+C\). It increases on \((0,\pi/2)\) and \((3\pi/2,2\pi)\), decreases on \((\pi/2,3\pi/2)\), has a local maximum at \(\pi/2\), and a local minimum at \(3\pi/2\). In part (g), the positive-to-negative change at \(-2\) gives a local maximum, and the negative-to-positive change at \(1\) gives a local minimum. In part (h), \(f\) is concave up on \((-3,0)\) and concave down on \((0,4)\); zero is an inflection input if the corresponding point of \(f\) exists. In part (i), all antiderivatives are \(f(x)=x^2+C\). The initial value gives \(C=3\), so \(f(x)=x^2+3\). In part (j), a horizontal tangent produces \(f'(c)=0\); a corner produces unequal finite one-sided derivative limits and no derivative value; a vertical tangent produces unbounded derivative behavior.