To analyze a function, you need to be able to connect the behavior of a graph with its derivatives. Many of the multiple-choice questions on the AP exam on this topic involve reasoning through these connections rather than computing any exact values.
Connecting graphs of f,f′, and f′′
Recall from the previous sections that
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 undefined, x=a is a critical point.
f(a) is an extrema if f′(x) changes sign around 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'
Determine which of the 4 graphs A through D could be the graph of f(x).
Choices for f(x)
First, locate the critical points of f(x), which occur when f′(x)=0 or undefined. Since the graph of f′(x) crosses the x-axis at x=−2,1, and 3, these are the critical points of f(x). We can classify them using the 1st derivative test:
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.
Or using the 2nd derivative test:
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 these extrema, so A and D can be eliminated.
However, the issue with graph B is that it’s made of straight line segments with sharp corners, so it can be expressed only as a piecewise function. Its derivative would just be constant values on each segment, with the slope undefined at the corner points x=−2,1, and 3, even though the graph of f′(x) clearly shows that f′(−2),f′(1), and f′(3) all equal 0.
So graph B can be eliminated, making graph C 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
Solution
(spoiler)
Answer:x=2
At x=−3,f(−3)<0 (f is below the x-axis) but it’s a relative min, which means 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 so f′(−1)>0. Since f(−1)<f′(−1), this answer choice can also be eliminated.
At x=1,f(1)>0 and it’s a relative max so f′(1)=0. However, because it’s concave down, f′′(1)<0. Since f′′(1)<f′(1)<f(1), choice x=1 can be eliminated.
At x=3,f(3)=0. The function is decreasing, so f′(3)<0. Because f(x) is concave up, f′′(3)>0. Since f′(3)<f(3)<f′′(3), this is the correct point.
Curve sketching
You may have sketched general shapes of polynomial, rational, exponential, and various other functions in Precalculus by locating points, intercepts, and determining end behavior, but with calculus, being able to find relative extrema and determining concavity allows you to sketch much more accurate and complex graphs. To sketch the shape of a function:
Find domain, x and y-intercepts, and asymptotes (if any)
Find critical points (1st derivative f′)
Test intervals for increasing/decreasing behavior of f and/or classify extrema
Find inflection points (2nd derivative f′′)
Test intervals for concavity of f
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
f′(x)=(x2−1)2(x2−1)(1)−(x)(2x)
=(x2−1)2x2−1−2x2
=(x2−1)2−x2−1
f′(x) is undefined when x=−1 and 1 (vertical asymptotes where f(x) is undefined). So not critical points.
f′(x)=0 when −x2−1=0→ no solutions. So there are no relative extrema.
Step 3: Sign of f′(x)
Plug into f′(x) a point in each region around the asymptotes x=−1 and x=1 that split up the domain. Here’s the sign diagram:
Sign diagram for f'(x)
Since the 1st derivative is negative, f(x) is decreasing in all 3 regions. You may find it visually helpful to draw arrows representing the behavior of f (negative slopes in the image above) and reminders such as open circles for where the original function f is undefined.
Step 4: 2nd derivative
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: f′′(x)=0 when x=0 only
Step 5: Determine concavity
Plug points in each region into f′′(x) to determine its sign. Vertical asymptotes x=−1 and 1 can’t be inflection points, but they’re important to include on the sign diagram for f′′ to sketch accurate curves.
Sign diagram for f''(x)
As shown, it can be helpful to draw a mini concave up or down curve depending on the sign of f′′.
Step 6: Putting it all together
Putting everything together on one sign diagram allows you to narrow down the shape of the curve - e.g. on the interval (1,∞),f(x) is decreasing and concave up, so it’s helpful to draw only that half of the concavity curve.
f, f', and f''
Plot the asymptotes and intercepts and then fill in the curves for the graph of f(x):
Final graph of f(x)
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