Premed · Premed · Calculus 1
Lecture 14: How Derivatives Affect Graph Shape
Calculus I — Differential Calculus
Learning Objectives
By the end of this lecture, students will be able to:
- Use the first derivative to determine where a function is increasing or decreasing
- Apply the First Derivative Test to classify local extrema
- Use the second derivative to determine concavity
- Find inflection points
- Apply the Second Derivative Test to classify local extrema
- Sketch curves using derivative information
Lecture Content
I. Increasing and Decreasing Functions (Review and Detail)
The Increasing/Decreasing Test, established as a corollary of the Mean Value Theorem, provides the foundation for analyzing graph shape. If f'(x) > 0 on an interval, then f is increasing on that interval; if f'(x) < 0, then f is decreasing. To determine the intervals of increase and decrease, find all critical numbers (where f'(x) = 0 or f'(x) does not exist), which divide the domain into intervals. Then test the sign of f'(x) in each interval using a test point. The function is increasing where f' is positive and decreasing where f' is negative.
II. The First Derivative Test
The First Derivative Test classifies critical numbers as local maxima, local minima, or neither by examining how the sign of f' changes. At a critical number c, if f' changes from positive to negative, then f has a local maximum at c. If f' changes from negative to positive, then f has a local minimum at c. If f' does not change sign at c, then f has neither a local maximum nor a local minimum there.
For example, consider f(x) = x^3 - 3x. The derivative is f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1), with critical numbers x = -1 and x = 1. A sign analysis reveals that f'(x) > 0 for x < -1, f'(x) < 0 for -1 < x < 1, and f'(x) > 0 for x > 1. At x = -1, f' changes from positive to negative, indicating a local maximum with f(-1) = 2. At x = 1, f' changes from negative to positive, indicating a local minimum with f(1) = -2.
<image>A sign chart and corresponding graph for f(x) = x^3 - 3x. Top: a number line showing critical points x = -1 and x = 1, with + signs on (-infinity, -1), - signs on (-1, 1), and + signs on (1, infinity). Arrows indicate increasing/decreasing. Bottom: the graph of the cubic function with the local maximum at (-1, 2) and local minimum at (1, -2) labeled with dots. Title: "First Derivative Test applied to f(x) = x^3 - 3x."</image>
III. Concavity
A function is concave up on an interval if its graph lies above all its tangent lines on that interval. The curve "opens upward" like a cup, and f'(x) is increasing. The test is that f''(x) > 0 implies the function is concave up. Conversely, a function is concave down on an interval if its graph lies below all its tangent lines. The curve "opens downward" like a cap, f'(x) is decreasing, and f''(x) < 0 on the interval.
IV. Inflection Points
A point where concavity changes -- from up to down or from down to up -- is called an inflection point. At an inflection point, the tangent line crosses the curve. To find inflection points, compute f''(x), find where f''(x) = 0 or f''(x) is undefined, and then check that f'' actually changes sign at these points, since a sign change is required.
It is important to note that f''(c) = 0 does not guarantee an inflection point. For example, f(x) = x^4 has f''(0) = 0, but x = 0 is not an inflection point because the concavity does not change there -- the function is concave up on both sides.
V. The Second Derivative Test
The Second Derivative Test offers an alternative to the First Derivative Test for classifying critical numbers. If c is a critical number where f'(c) = 0, then: if f''(c) > 0, the function has a local minimum at c (the graph is concave up, like a valley); if f''(c) < 0, the function has a local maximum at c (the graph is concave down, like a hilltop); and if f''(c) = 0, the test is inconclusive and the First Derivative Test must be used instead. The advantage of this test is its speed when f'' is easy to compute, but its limitation is that it fails precisely when f''(c) = 0.
<image>Two panels illustrating the Second Derivative Test. Panel A: A concave-up curve (f'' > 0) with a critical point at the bottom — labeled "local minimum, f''(c) > 0." The tangent line at the minimum is horizontal, and the curve lies above it. Panel B: A concave-down curve (f'' < 0) with a critical point at the top — labeled "local maximum, f''(c) < 0." The tangent line at the maximum is horizontal, and the curve lies below it. Title: "Second Derivative Test: concavity determines the type of extremum."</image>
VI. Curve Sketching Procedure
A systematic approach to graphing y = f(x) proceeds through seven steps. First, determine the domain of f. Second, find the intercepts: x-intercepts by setting f(x) = 0, and the y-intercept by evaluating f(0). Third, check for symmetry -- is f even, odd, or periodic? Fourth, find asymptotes: vertical asymptotes where the denominator is zero, horizontal asymptotes by evaluating limits at infinity, and oblique asymptotes if applicable. Fifth, perform a first derivative analysis to find critical numbers, determine intervals of increase and decrease, and locate local extrema. Sixth, perform a second derivative analysis to find intervals of concavity and inflection points. Seventh, plot key points and sketch the curve, using all of the information gathered.
VII. Complete Curve Sketching Example
Consider f(x) = x^4 - 4x^3. The domain is all real numbers. The intercepts are found from f(x) = x^3(x - 4) = 0, giving x = 0 and x = 4, with y-intercept f(0) = 0.
The first derivative is f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3), with critical numbers x = 0 and x = 3. The sign analysis shows f' < 0 for x < 0 (decreasing), f' < 0 for 0 < x < 3 (still decreasing), and f' > 0 for x > 3 (increasing). At x = 0, there is no sign change, so it is neither a maximum nor a minimum. At x = 3, f' changes from negative to positive, giving a local minimum with f(3) = 81 - 108 = -27.
The second derivative is f''(x) = 12x^2 - 24x = 12x(x - 2), which equals zero at x = 0 and x = 2. The concavity is up for x < 0, down for 0 < x < 2, and up again for x > 2. Inflection points occur at x = 0 (where f(0) = 0) and x = 2 (where f(2) = 16 - 32 = -16).
<image>The complete graph of f(x) = x^4 - 4x^3 with all key features labeled: x-intercepts at (0, 0) and (4, 0), the local minimum at (3, -27), inflection points at (0, 0) and (2, -16). Regions of concave up and concave down are indicated with annotations. Arrows show increasing/decreasing behavior. The sign charts for f'(x) and f''(x) are displayed above the graph as number lines. Title: "Complete curve sketch of f(x) = x^4 - 4x^3."</image>
