Concavity and inflection points
A graph can rise while bending up or while bending down. How does the second derivative tell which way a graph bends, where the bend reverses, and whether a tangent line runs above or below the curve?
Where you are 11 in Applications of derivatives
Why this exists
Concavity: when the rate of change itself changes described the bend of a graph with secant slopes and second differences, and Linear approximation and differentials decided whether a tangent-line estimate ran high or low by looking at a picture. Both questions ask whether the slope f' is increasing, and whether a function increases is read from its derivative. So the answer sits in the Second and higher derivatives. This lesson turns the bend of a graph into a sign of f'' computed from a formula, finds where the bend reverses, and settles which way a tangent-line estimate errs without a picture.
Definition
Day 1: concave up, concave down, inflection points
Let f be differentiable on an interval I. Then f is concave up on I if f' is increasing on I, and concave down on I if f' is decreasing on I.
This lesson takes the slope description as the definition. The other description, that the graph lies above each of its tangent lines on I (concave up) or below them (concave down), follows from it and implies it back, as proved under “Why it is true”. The secant form of Concavity: when the rate of change itself changes agrees too. Concavity is reported here on open intervals.
A point of inflection is a point (c, f(c)) of the graph such that f is continuous at c and the concavity changes at c: f is concave up on an open interval ending at c and concave down on one starting at c, or the reverse. Nothing is asked of f' or f'' at c itself.
Day 2: estimates and rates
The linearization of f at a is L(x) = f(a) + f'(a)(x - a), as in Linear approximation and differentials. The estimate L(x) is an overestimate of f(x) when L(x) > f(x) and an underestimate when L(x) < f(x).
When f(t) is a quantity changing in time, f''(t) is the rate at which its rate f'(t) changes. A quantity can grow (f' > 0) while its growth slows (f'' < 0).
Intuition
Read a concave up graph from left to right and its tangent lines turn counterclockwise, each steeper than the one before. A curve that keeps turning upward bends away from each tangent toward the top, so it sits above its tangents. A concave down graph turns clockwise and sits below them. At an inflection point the turning reverses, and the tangent there crosses the graph (Figure 1).
Rising and bending are separate facts: the sign of f' says whether the graph goes up, the sign of f'' whether its slope is growing.
Results
The concavity test. Suppose f'' exists at every point of an open interval I.
- If f''(x) > 0 for every x in I, then f is concave up on I.
- If f''(x) < 0 for every x in I, then f is concave down on I.
If f' is also continuous at an end of I, that end may be included, as in Increasing and decreasing from the derivative.
Tangent lines. If f is concave up on an interval I and a is in I, then
The graph lies above its tangent at a, so the linearization at a is an underestimate. If f is concave down on I, the inequality reverses and the linearization is an overestimate. Conversely, if a differentiable function lies above each of its tangent lines on I, it is concave up there.
Where inflection points can be. If (c, f(c)) is a point of inflection, then f''(c) = 0 or f''(c) does not exist. Neither condition guarantees an inflection point:
- f(x) = x^4 has f''(0) = 0, but f''(x) = 12x^2 > 0 for x \ne 0 and f'(x) = 4x^3 is increasing on (-\infty, \infty). It is concave up everywhere, with no inflection point.
- f(x) = x^{1/3} has no second derivative at 0, yet \displaystyle f''(x) = -\frac{2}{9}x^{-5/3} is positive for x < 0 and negative for x > 0. Since f is continuous at 0, (0, 0) is a point of inflection.
So f''(c) = 0 is neither sufficient nor necessary. The candidates go on a sign chart for f'': mark the inputs where f'' = 0, where f'' does not exist, and where f is undefined; then one test value in each piece fixes the sign of f'' there. An inflection point sits at a marked input in the domain of f where the sign changes.
Why it is true
The concavity test is the increasing test of Increasing and decreasing from the derivative applied to f' instead of f. Since f'' exists on I, f' is differentiable on I, and its derivative f'' is positive there, so f' is increasing on I by the mean value theorem (The mean value theorem).
Tangent lines. Let f' be increasing on I, with a in I. For x > a in I, the mean value theorem gives a c with a < c < x and f(x) - f(a) = f'(c)(x - a). Since c > a, f'(c) > f'(a), and x - a > 0, so f(x) - f(a) > f'(a)(x - a). For x < a, the theorem gives c with x < c < a and f(a) - f(x) = f'(c)(a - x); now f'(c) < f'(a) and a - x > 0, so f(a) - f(x) < f'(a)(a - x), which rearranges to the same inequality.
For the converse, suppose the graph lies above every tangent line, and take a < b in I. The tangents at a and at b give
Adding and cancelling f(a) + f(b) leaves 0 > (b - a)\big(f'(a) - f'(b)\big), so f'(a) < f'(b): f' is increasing. The concave down case reverses every inequality.
Secants. If f' is increasing and p < q < r lie in I, the mean value theorem writes the secant slopes as m(p, q) = f'(c_1) and m(q, r) = f'(c_2) with c_1 < q < c_2, so m(p, q) < m(q, r). (The converse is stated without proof.)
Inflection candidates. Suppose the concavity changes at c, say concave up on (c - \delta, c) and concave down on (c, c + \delta), and suppose f''(c) exists. Then f' is differentiable, hence continuous, at c. It increases on (c - \delta, c) and decreases on (c, c + \delta), so f'(c) is the largest value of f' near c: a local maximum of f' at an interior point. By Fermat’s theorem in Critical points and local versus absolute extrema, the derivative of f' is 0 there, that is, f''(c) = 0. The reverse change gives a local minimum of f' and the same conclusion.
Worked examples
Day 1: Talia’s quartic
Talia wants the intervals of concavity and the inflection points of f(x) = x^4 - 2x^3 - 12x^2 + 5x.
SolutionA polynomial has derivatives of every order, so the only split points are zeros of f'':
The zeros -1 and 2 cut the line into three pieces where f'' is continuous and never 0, so one test value gives each sign:
| Piece | Test | f'' | Sign |
|---|---|---|---|
| (-\infty, -1) | -2 | 48 | + |
| (-1, 2) | 0 | -24 | - |
| (2, \infty) | 3 | 48 | + |
By the concavity test, f is concave up on (-\infty, -1) and on (2, \infty), and concave down on (-1, 2). The sign changes at both split points and f is continuous there, so both are inflection points. With f(-1) = 1 + 2 - 12 - 5 = -14 and f(2) = 16 - 16 - 48 + 10 = -38, the inflection points are (-1, -14) and (2, -38). The zeros of f' played no part.
Day 1: Nalesoni’s cube root
Nalesoni studies g(x) = x^{1/3}(x - 8). Find where g is concave up and concave down, and its inflection points.
SolutionWrite g(x) = x^{4/3} - 8x^{1/3}. For x \ne 0 the power rule gives
and differentiating again,
The last form uses x^{-2/3} = x \cdot x^{-5/3}. As x \to 0, g'(x) \to -\infty, so neither g'(0) nor g''(0) exists. The split points are -4, where g'' = 0, and 0, where g'' does not exist. Since x^{5/3} = \big(x^{1/3}\big)^5 has the sign of x, the test values give \displaystyle g''(-8) = \frac{4(-4)}{9(-32)} = \frac{1}{18} > 0, \displaystyle g''(-1) = \frac{4(3)}{9(-1)} = -\frac{4}{3} < 0 and \displaystyle g''(1) = \frac{20}{9} > 0.
So g is concave up on (-\infty, -4), concave down on (-4, 0) and concave up on (0, \infty). The sign changes at -4 and at 0, and g is continuous at both, so the inflection points are (-4, g(-4)) and (0, 0), where \displaystyle g(-4) = (-4)^{1/3}(-12) = 12\sqrt[3]{4} \approx 19.05. At (0, 0) the graph has a vertical tangent.
Day 1: Viliami’s zero that does not count
Viliami finds that f(x) = 3x^5 - 5x^4 has f''(0) = 0 and calls (0, 0) an inflection point. Find the concavity of f and decide.
Solutionf'(x) = 15x^4 - 20x^3 and f''(x) = 60x^3 - 60x^2 = 60x^2(x - 1), with zeros 0 and 1. Test values: f''(-1) = -120, \displaystyle f''\left(\frac{1}{2}\right) = -\frac{15}{2}, f''(2) = 240.
The factor x^2 keeps the sign negative on both sides of 0. So f' is decreasing on (-\infty, 0] and on [0, 1], and these join, just as increasing intervals join in Increasing and decreasing from the derivative: f is concave down on (-\infty, 1) and concave up on (1, \infty). The only inflection point is (1, f(1)) = (1, -2).
Viliami is not right: f''(0) = 0, but the concavity does not change at 0. (Since f'(0) = 0, 0 is a critical point; sorting those is the work of The first and second derivative tests.)
Day 1: Talia’s cubic built to order
Find a and b so that f(x) = ax^3 + bx^2 + 4x has a point of inflection at (1, 6).
Solutionf''(x) = 6ax + 2b exists everywhere, so an inflection point at x = 1 forces f''(1) = 0: 6a + 2b = 0, that is, b = -3a. The point must be on the graph: f(1) = a + b + 4 = 6, so a + b = 2. Substituting, a - 3a = 2, so a = -1 and b = 3.
f''(1) = 0 is only necessary, so check the change. With f(x) = -x^3 + 3x^2 + 4x, f''(x) = -6x + 6 = -6(x - 1), positive for x < 1 and negative for x > 1. The concavity changes, and (1, 6) is an inflection point.
Day 2: Viliami’s two linearizations
(a) Use the linearization of \ln x at 1 to estimate \ln 1.08 and \ln 0.95. (b) Use the linearization of \tan x at 0 to estimate \tan 0.1 and \tan(-0.1). Decide from f'' whether each estimate is over or under.
Solution(a) For f(x) = \ln x, \displaystyle f'(x) = \frac{1}{x} and \displaystyle f''(x) = -\frac{1}{x^2} < 0 on (0, \infty), so f is concave down on an interval containing 0.95, 1 and 1.08. With f(1) = 0 and f'(1) = 1, L(x) = x - 1, giving \ln 1.08 \approx 0.08 and \ln 0.95 \approx -0.05. A concave down graph lies below its tangent, so both are overestimates. Indeed \ln 1.08 \approx 0.07696 and \ln 0.95 \approx -0.05129.
(b) For f(x) = \tan x, f'(x) = \sec^2 x and, by the chain rule, f''(x) = 2\sec x \cdot \sec x \tan x = 2\sec^2 x \tan x. Since \sec^2 x > 0, f'' has the sign of \tan x: negative on \displaystyle \left(-\frac{\pi}{2}, 0\right) and positive on \displaystyle \left(0, \frac{\pi}{2}\right). So (0, 0) is an inflection point and L(x) = x. On \displaystyle \left[0, \frac{\pi}{2}\right), f is concave up (f' is continuous at 0, so the end is included), so \tan 0.1 \approx 0.1 is an underestimate. On \displaystyle \left(-\frac{\pi}{2}, 0\right], f is concave down, so \tan(-0.1) \approx -0.1 is an overestimate. Indeed \tan 0.1 \approx 0.10033. A tangent at an inflection point crosses the curve, so it errs in opposite directions on its two sides.
Day 2: Nalesoni’s plant
Nalesoni’s plant is h(t) = 5 + 0.9t^2 - 0.05t^3 centimeters tall t weeks after planting, for 0 \le t \le 12. (a) Show the plant grows the whole time, and find where h is concave up and down. What does that mean for the plant? (b) When is it growing fastest, and how fast? (c) Compare weeks 3 and 9. (d) Estimate h(7) from the tangent line at t = 6, and say whether the estimate is over or under.
Solution(a) h'(t) = 1.8t - 0.15t^2 = 0.15t(12 - t), which is positive on (0, 12), so h is increasing on [0, 12]. Next, h''(t) = 1.8 - 0.3t = 0.3(6 - t), positive on (0, 6) and negative on (6, 12). So h is concave up on (0, 6) and concave down on (6, 12), with an inflection point at t = 6. For the first six weeks the growth rate increases; after that the plant still grows, but each week more slowly.
(b) h' is increasing on [0, 6] and decreasing on [6, 12], so the growth rate is largest at the inflection point: h'(6) = 0.15(6)(6) = 5.4 centimeters per week, when the plant is h(6) = 5 + 32.4 - 10.8 = 26.6 centimeters tall.
(c) h'(3) = 0.15(3)(9) = 4.05 and h'(9) = 0.15(9)(3) = 4.05 centimeters per week: the same rate. But h''(3) = 0.9 and h''(9) = -0.9 centimeters per week per week. At week 3 the growth is speeding up; at week 9 it is slowing down.
(d) L(t) = 26.6 + 5.4(t - 6), so h(7) \approx 32.0. Since h'' < 0 on (6, 12) and h' is continuous at 6, h' is decreasing on [6, 12], which contains 6 and 7, so the estimate is over. Check: h(7) = 5 + 44.1 - 17.15 = 31.95, so it is over by 0.05 centimeter.
Day 2: reading concavity from the graph of f’
Figure 3 is the graph of f' on [-1, 5]. It has a level tangent at (1, 1) and touches the axis at 3. (a) Where is f concave up and concave down? (b) Find the x-coordinates of the inflection points of f. (c) Viliami says x = 0 is an inflection point, because the graph crosses the axis there. Is he right? (d) What happens on the graph of f at x = 3?
(a) Concavity asks whether f' is increasing, so read whether this graph rises or falls, not where it sits. It rises on (-1, 1), falls on (1, 3) and rises on (3, 5). So f is concave up on (-1, 1) and (3, 5) and concave down on (1, 3).
(b) The concavity changes at x = 1 and x = 3, the peak and the valley of the graph of f'. These are the inflection points’ x-coordinates.
(c) No. Where f' crosses the axis, f changes direction, not bend. The graph of f' rises through 0, so f is concave up on both sides.
(d) f'(3) = 0, so f has a horizontal tangent there, and the concavity changes. Since f' \ge 0 on both sides, f flattens out and climbs again, like x^3 at 0. The graph was drawn from \displaystyle f'(x) = \frac{1}{4}x(x - 3)^2, and its derivative \displaystyle f''(x) = \frac{3}{4}(x - 1)(x - 3) changes sign at exactly 1 and 3, as read.
Boundaries
- The test reads f'' only where it exists. The bend can reverse where f'' does not exist (x^{1/3}, Example 2).
- A zero of f'' is only a candidate. x^4 and Example 3 have f'' = 0 with no change of concavity.
- The test goes one way. Concave up does not force f'' > 0 everywhere: x^4 has f''(0) = 0.
- An inflection point needs c in the domain and f continuous there. \dfrac{1}{x} changes concavity at 0 but has no inflection point.
- The tangent-line comparison needs one concavity between a and x. Across an inflection point the estimate may err either way.
- Telling a peak from a valley by the sign of f'' is the second derivative test, in The first and second derivative tests, not here.
Where this goes next
The sign of f'' at a critical point gives the second derivative test in The first and second derivative tests. Signs of f' and f'' together build a whole graph in Curve sketching and connecting f, f' and f''. Velocity and acceleration of the same sign mean speeding up, in Position, velocity and acceleration. A logistic population grows fastest at its inflection point (Logistic growth). Concavity decides which way an approximation leans in Euler's method and The trapezoidal rule and Simpson's rule, as it does for a tangent line here.
Exercises
Day 1: concavity and inflection points from a formula
For Exercises 1 to 8, find the open intervals on which f is concave up and concave down, and every point of inflection.
- f(x) = x^3 - 6x^2 + 5x - 2
- f(x) = x^4 + 4x^3 - 18x^2 + 7
- f(x) = x^4 - 8x^3 + 24x^2
- f(x) = (x - 1)e^{-x}
- f(x) = \ln(x^2 + 9)
- f(x) = (x - 1)^{1/3} + x
- f(x) = e^x \sin x on [0, 2\pi]
- f(x) = \dfrac{x^2 + 1}{x - 2}
- Find a and b so that f(x) = x^3 + ax^2 + bx has a point of inflection at (2, -4).
Day 1: reasoning
- Prove that every cubic f(x) = ax^3 + bx^2 + cx + d with a \ne 0 has exactly one point of inflection, at \displaystyle x = -\frac{b}{3a}.
- For f(x) = (x - 1)^4 + x, Talia notes f''(1) = 0 and concludes that (1, 1) is a point of inflection. Is she right? Explain.
- Viliami claims that a function decreasing on (-\infty, \infty) cannot be concave up on (-\infty, \infty). Give a counterexample and describe its graph in words.
Day 2: tangent-line estimates
- Use the linearization of \displaystyle \sqrt{x} at 25 to estimate \displaystyle \sqrt{26} and \displaystyle \sqrt{24}. Decide from f'' whether each is over or under, then check.
- Use the linearization of x^3 at 1 to estimate 0.97^3 and 1.03^3. Over or under?
- Use the linearization of \dfrac{1}{x} at 2 to estimate \dfrac{1}{2.1} and \dfrac{1}{1.9}. Over or under?
- Let f(x) = x^3 + 3x^2 - 2. Show that (-1, 0) is an inflection point, find the linearization L at -1, show that f(x) - L(x) = (x + 1)^3, and say whether L(-0.9) and L(-1.1) are over or under.
Day 2: concavity in context and from graphs
- A drink Talia pours cools so that its temperature is T(t) = 70 + 110e^{-0.1t} degrees Fahrenheit t minutes after it is poured. (a) Find the signs of T' and T'' and say what they mean. (b) Estimate T(11) from the tangent line at t = 10. Over or under?
- Give the signs of f' and f'' for each quantity f(t). (a) The number of subscribers is still rising, but more slowly each month. (b) A patient’s temperature is falling, and falling faster each hour. (c) Gasoline prices fell this year, but the decline is leveling off.
- Figure 4 is the graph of g'' on [-3, 4]. On which open intervals is g concave up, and on which concave down?
- Using Figure 4, find the x-coordinates of the inflection points of g. Why is x = 1 not one?
- A function h has h' increasing on (-\infty, 2) and decreasing on (2, \infty), with h(2) = 1 and h'(2) = 5. (a) Describe the concavity of h and its inflection point. (b) Find the linearization at 2, and say whether its values at 2.1 and 1.9 are over or under.
Day 2: reasoning
- Prove that \sin x < x for 0 < x < \pi.
- Nalesoni says: “In Figure 3 the graph of f' is above the axis on (0, 5), so f is concave up there.” Explain what is wrong, and what the height of the graph of f' does tell.
- Suppose f' is increasing on [c - \delta, c] and decreasing on [c, c + \delta]. Prove that the graph of f lies above its tangent line at c just left of c and below it just right of c.
- Suppose f''(x) > 0 for every x, and f(a) = f(b) = 0 with a < b. Prove that f(x) < 0 for every x between a and b.
Answers
- f''(x) = 6x - 12 = 6(x - 2). Concave down on (-\infty, 2), up on (2, \infty); inflection point (2, -8).
- f''(x) = 12x^2 + 24x - 36 = 12(x + 3)(x - 1), signs +, -, +. Concave up on (-\infty, -3) and (1, \infty), down on (-3, 1); inflection points (-3, -182) and (1, -6).
- f''(x) = 12x^2 - 48x + 48 = 12(x - 2)^2, positive except at 2. f' is increasing on (-\infty, 2] and [2, \infty), which join: concave up on (-\infty, \infty), no inflection point.
- f'(x) = (2 - x)e^{-x} and f''(x) = (x - 3)e^{-x}, with e^{-x} > 0. Concave down on (-\infty, 3), up on (3, \infty); inflection point (3, 2e^{-3}).
- f'(x) = \dfrac{2x}{x^2 + 9} and f''(x) = \dfrac{2(9 - x^2)}{(x^2 + 9)^2}. Concave down on (-\infty, -3) and (3, \infty), up on (-3, 3); inflection points (-3, \ln 18) and (3, \ln 18).
- For x \ne 1, f'(x) = \dfrac{1}{3}(x - 1)^{-2/3} + 1 and f''(x) = -\dfrac{2}{9}(x - 1)^{-5/3}, which has the sign of 1 - x. Concave up on (-\infty, 1), down on (1, \infty); f is continuous at 1, so (1, 1) is an inflection point, though f''(1) does not exist.
- f'(x) = e^x(\sin x + \cos x) and f''(x) = 2e^x\cos x, zero at \displaystyle \frac{\pi}{2} and \displaystyle \frac{3\pi}{2}. Concave up on \displaystyle \left(0, \frac{\pi}{2}\right) and \displaystyle \left(\frac{3\pi}{2}, 2\pi\right), down on \displaystyle \left(\frac{\pi}{2}, \frac{3\pi}{2}\right); inflection points \displaystyle \left(\frac{\pi}{2}, e^{\pi/2}\right) and \displaystyle \left(\frac{3\pi}{2}, -e^{3\pi/2}\right).
- Division gives f(x) = x + 2 + \dfrac{5}{x - 2}, so f''(x) = \dfrac{10}{(x - 2)^3}. Concave down on (-\infty, 2), up on (2, \infty). No inflection point: 2 is not in the domain.
- a = -6, b = 6. f''(2) = 12 + 2a = 0 gives a = -6; f(2) = 8 - 24 + 2b = -4 gives b = 6. Then f''(x) = 6x - 12 changes sign at 2.
- f''(x) = 6ax + 2b exists everywhere and is 0 only at \displaystyle x = -\frac{b}{3a}, so that is the only candidate. It is a linear function with slope 6a \ne 0, so it changes sign there, and f is continuous: exactly one inflection point.
- No. f''(x) = 12(x - 1)^2 \ge 0, and f'(x) = 4(x - 1)^3 + 1 is increasing on (-\infty, \infty), so f is concave up everywhere. (1, 1) is on the graph, but the concavity does not change there.
- f(x) = e^{-x}: f'(x) = -e^{-x} < 0 and f''(x) = e^{-x} > 0. The graph falls everywhere, more and more slowly, flattening toward the x-axis.
- Over, both. \displaystyle L(x) = 5 + \frac{1}{10}(x - 25) gives \displaystyle \sqrt{26} \approx 5.1 and \displaystyle \sqrt{24} \approx 4.9. \displaystyle f''(x) = -\frac{1}{4}x^{-3/2} < 0, so concave down on (0, \infty). Check: \displaystyle \sqrt{26} \approx 5.0990, \displaystyle \sqrt{24} \approx 4.8990.
- Under, both. L(x) = 1 + 3(x - 1) gives 0.91 and 1.09; f''(x) = 6x > 0 on (0, \infty). Check: 0.97^3 = 0.912673, 1.03^3 = 1.092727.
- Under, both. \displaystyle L(x) = \frac{1}{2} - \frac{1}{4}(x - 2) gives 0.475 and 0.525; \displaystyle f''(x) = \frac{2}{x^3} > 0 on (0, \infty). Check: \displaystyle \frac{1}{2.1} \approx 0.47619, \displaystyle \frac{1}{1.9} \approx 0.52632.
- f''(x) = 6x + 6 changes sign at -1 and f(-1) = 0, so (-1, 0) is an inflection point. f'(-1) = 3 - 6 = -3, so L(x) = -3(x + 1), and f(x) - L(x) = x^3 + 3x^2 + 3x + 1 = (x + 1)^3. At -0.9 this is 0.001 > 0, so L(-0.9) = -0.3 is under; at -1.1 it is -0.001, so L(-1.1) = 0.3 is over.
- (a) T'(t) = -11e^{-0.1t} < 0 and T''(t) = 1.1e^{-0.1t} > 0: the drink cools, more and more slowly. (b) T(10) = 70 + 110e^{-1} \approx 110.47 and T'(10) = -11e^{-1} \approx -4.05, so T(11) \approx 106.42. Concave up, so under; indeed T(11) = 70 + 110e^{-1.1} \approx 106.62.
- (a) f' > 0, f'' < 0. (b) f' < 0, f'' < 0. (c) f' < 0, f'' > 0.
- g'' > 0 on (-3, -2) and (3, 4), so concave up there; g'' < 0 on (-2, 1) and (1, 3), and since g' is continuous at 1 these join: concave down on (-2, 3).
- x = -2 and x = 3, where g'' changes sign. At 1, g'' touches 0 but is negative on both sides, so the concavity does not change.
- (a) Concave up on (-\infty, 2), down on (2, \infty); inflection point (2, 1). (b) L(x) = 1 + 5(x - 2). L(2.1) = 1.5 is over (concave down on [2, \infty)); L(1.9) = 0.5 is under (concave up on (-\infty, 2]).
- f(x) = \sin x has f''(x) = -\sin x < 0 on (0, \pi), and f'(x) = \cos x is continuous at 0, so f is concave down on [0, \pi). Its tangent at 0 is y = x, so by the tangent-line result \sin x < x for 0 < x < \pi.
- The height of f' gives the sign of the slope of f: f is increasing on [0, 5]. Concavity depends on whether f' is increasing. On (1, 3) the graph of f' falls, so f is concave down there although f' > 0.
- On [c - \delta, c], f' is increasing, so by the tangent-line result with a = c at the right end, f(x) > f(c) + f'(c)(x - c) for c - \delta \le x < c. On [c, c + \delta], f' is decreasing, so the inequality reverses for c < x \le c + \delta.
- Rolle’s theorem gives c in (a, b) with f'(c) = 0. Since f'' > 0, f' is increasing: f' < 0 on (a, c) and f' > 0 on (c, b). So f is decreasing on [a, c], giving f(x) < f(a) = 0 for a < x \le c, and increasing on [c, b], giving f(x) < f(b) = 0 for c \le x < b.