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248 derivatives and graphs
3. 4 f ′′^ and the Shape of f
The first derivative of a function provides information about the shape of the function, so the second derivative of a function provides informa- tion about the shape of the first derivative, which in turn will provide additional information about the shape of the original function f. In this section we investigate how to use the second derivative (and the shape of the first derivative) to reach conclusions about the shape of the original function. The first derivative tells us whether the graph of f is increasing or decreasing. The second derivative will tell us about the “concavity” of f : whether f is curving upward or downward.
Concavity
Graphically, a function is concave up if its graph is curved with the opening upward (see margin); similarly, a function is concave down if its graph opens downward. The concavity of a function can be impor- tant in applied problems and can even affect billion-dollar decisions. An Epidemic : Suppose you, as an official at the CDC, must de- cide whether current methods are effectively fighting the spread of a disease—or whether more drastic measures are required. In the margin figure, f (x) represents the number of people infected with the disease at time x in two different situations. In both cases the number of people with the disease, f (now), and the rate at which new people are getting sick, f ′(now), are the same. The difference is the concavity of f , and that difference might have a big effect on your decision. In (a), f is concave down at “now,” and it appears that the current methods are starting to bring the epidemic under control; in (b), f is concave up, and it appears that the epidemic is growing out of control. Usually it is easy to determine the concavity of a function by exam- ining its graph, but we also need a definition that does not require a graph of the function, a definition we can apply to a function described by a formula alone.
Definition : Let f be a differentiable function.
- f is concave up at a if the graph of f is above the tangent line L to f for all x close to (but not equal to) a:
f (x) > L(x) = f (a) + f ′(a)(x − a)
- f is concave down at a if the graph of f is below the tangent line L to f for all x close to (but not equal to) a:
f (x) < L(x) = f (a) + f ′(a)(x − a)
- 4 f ′′^ and the shape of f 249
The margin figure shows the concavity of a function at several points. The next theorem provides an easily applied test for the concavity of a function given by a formula.
The Second Derivative Condition for Concavity : Let f be a twice differentiable function on an interval I.
(a) f ′′(x) > 0 on I ⇒ f ′(x) increasing on I ⇒ f concave up on I
(b) f ′′(x) < 0 on I ⇒ f ′(x) decreasing on I ⇒ f concave down on I
(c) f ′′(a) = 0 ⇒ no information ( f (x) may be concave up or concave down or neither at a)
Proof. (a) Assume that f ′′(x) > 0 for all x in I, and let a be any point in I. We want to show that f is concave up at a, so we need to prove that the graph of f (see margin) is above the tangent line to f at a: f (x) > L(x) = f (a) + f ′(a)(x − a) for x close to a. Assume that x is in I and apply the Mean Value Theorem to f on the interval with endpoints a and x: there is a number c between a and x so that
f ′(c) = f^ (x)^ −^ f^ (a) x − a ⇒ f (x) = f (a) + f ′(c)(x − a)
Because f ′′^ > 0 on I, we know that f ′′^ > 0 between a and x, so the Second Shape Theorem tells us that f ′^ is increasing between a and x. We will consider two cases: x > a and x < a.
- If x > a then x − a > 0 and c is in the interval [a, x] so a < c. Because f ′^ is increasing, a < c ⇒ f ′(a) < f ′(c). Multiplying each side of this last inequality by the positive quantity x − a yields f ′(a)(x − a) < f ′(c)(x − a). Adding f (a) to each side of this last inequality, we have:
L(x) = f (a) + f ′(a)(x − a) < f (a) + f ′(c)(x − a) = f (x)
- If x < a then x − a < 0 and c is in the interval [x, a] so c < a. Because f ′^ is increasing, c < a ⇒ f ′(c) < f ′(a). Multiplying each side of this last inequality by the negative quantity x − a yields f ′(c)(x − a) > f ′(a)(x − a) so:
f (x) = f (a) + f ′(c)(x − a) > f (a) + f ′(a)(x − a) = L(x)
In each case we see that f (x) is above the tangent line L(x).
(b) The proof of is similar.
- 4 f ′′^ and the shape of f 251
Solution. v′(t) = a(t) = 30 ⇒ v(t) = 30 t + K for some constant K. We also know v( 0 ) = 0 so 30( 0 ) + K = 0 ⇒ K = 0 and this v(t) = 30 t. Similarly, h′(t) = v(t) = 30 t ⇒ h(t) = 15 t^2 + C for some constant C. We know that h( 0 ) = 25 so 15 ( 0 )^2 + C = 25 ⇒ C = 25. Thus h(t) = 15 t^2 + 25 so h( 6 ) = 15 ( 6 )^2 + 25 = 565 m. J
f ′′^ and Extreme Values of f
The concavity of a function can also help us determine whether a critical point is a maximum or minimum or neither. For example, if a point is at the bottom of a concave-up function then that point is a minimum.
The Second Derivative Test for Extremes : Let f be a twice differentiable function.
(a) If f ′(c) = 0 and f ′′(c) < 0 then f is concave down and has a local maximum at x = c.
(b) If f ′(c) = 0 and f ′′(c) > 0 then f is concave up and has a local minimum at x = c.
(c) If f ′(c) = 0 and f ′′(c) = 0 then f may have a local maximum, a local minimum or neither at x = c.
Proof. (a) Assume that f ′(c) = 0. If f ′′(c) < 0 then f is concave down at x = c so the graph of f will be below the tangent line L(x) for values of x near c. The tangent line, however, is given by L(x) = f (c) + f ′(c)(x − c) = f (c) + 0 (x − c) = f (c), so if x is close to c then f (x) < L(x) = f (c) and f has a local maximum at x = c.
(b) The proof for a local minimum of f is similar.
(c) If f ′(c) = 0 and f ′′(c) = 0 , then we cannot immediately conclude anything about local maximums or minimums of f at x = c. The functions f (x) = x^4 , g(x) = −x^4 and h(x) = x^3 all have their first and second derivatives equal to zero at x = 0 , but f has a local minimum at 0 , g has a local maximum at 0 , and h has neither a local maximum nor a local minimum at x = 0.
The Second Derivative Test for Extremes is very useful when f ′′^ is easy to calculate and evaluate. Sometimes, however, the First Derivative Test—or simply a graph of the function—provides an easier way to determine if the function has a local maximum or a local minimum: it depends on the function and on which tools you have available. Practice 2. f (x) = 2 x^3 − 15 x^2 + 24 x − 7 has critical numbers x = 1 and x = 4. Use the Second Derivative Test for Extremes to determine whether f ( 1 ) and f ( 4 ) are maximums or minimums or neither.
252 derivatives and graphs
Inflection Points
Maximums and minimums typically occur at places where the second derivative of a function is positive or negative, but the places where the second derivative is 0 are also of interest.
Definition : An inflection point is a point on the graph of a function where the concavity of the function changes, from concave up to concave down or from concave down to concave up.
Practice 3. Which of the labeled points in the margin figure are inflec- tion points? To find the inflection points of a function we can use the second derivative of the function. If f ′′(x) > 0 , then the graph of f is concave up at the point (x, f (x)) so (x, f (x)) is not an inflection point. Similarly, if f ′′(x) < 0 then the graph of f is concave down at the point (x, f (x)) and the point is not an inflection point. The only points left that can possibly be inflection points are the places where f ′′(x) = 0 or where f ′′(x) does not exits (in other words, where f ′^ is not differentiable). To find the inflection points of a function we only need check the points where f ′′(x) is 0 or undefined. If f ′′(c) = 0 or is undefined, then the point (c, f (c)) may or may not be an inflection point—we need to check the concavity of f on each side of x = c. The functions in the next example illustrate what can happen at such a point. Example 2. Let f (x) = x^3 , g(x) = x^4 and h(x) = 3
x (see margin). For which of these functions is the point (0, 0) an inflection point?
Solution. Graphically, it is clear that the concavity of f (x) = x^3 and h(x) = 3
x changes at (0, 0), so (0, 0) is an inflection point for f and h. The function g(x) = x^4 is concave up everywhere, so (0, 0) is not an inflection point of g. f (x) = x^3 ⇒ f ′(x) = 3 x^2 ⇒ f ′′(x) = 6 x so the only point at which f ′′(x) = 0 or is undefined ( f ′^ is not differentiable) is at x = 0. If x < 0 then f ′′(x) < 0 so f is concave down; if x > 0 then f ′′(x) > 0 so f is concave up. Thus at x = 0 the concavity of f changes so the point (0, f ( 0 )) = (0, 0) is an inflection point of f (x) = x^3. g(x) = x^4 ⇒ g′(x) = 4 x^3 ⇒ g′′(x) = 12 x^2 so the only point at which g′′(x) = 0 or is undefined is at x = 0. But g′′(x) > 0 (so g is concave up) for any x 6 = 0. Thus the concavity of g never changes, so the point (0, g( 0 )) = (0, 0) is not an inflection point of g(x) = x^4. h(x) = 3
x = x (^13) ⇒ h′(x) =
x−^ (^23) ⇒ h′′(x) = −
x−^ (^59) so h′′^ is not defined if x = 0 (and h′′(x) 6 = 0 elsewhere); h′′(negative number) > 0
254 derivatives and graphs
At which values of x labeled in the figure below is the point (x, f (x)) an inflection point?
At which values of x labeled in the figure below is the point (x, g(x)) an inflection point?
How many inflection points can a:
(a) quadratic polynomial have? (b) cubic polynomial have? (c) polynomial of degree n have?
Fill in the table with “+,” “−,”or “ 0 ” for the func- tion graphed below. x f (x) f ′(x) f ′′(x) 0 1 2 3 4
Fill in the table with “+,” “−,”or “ 0 ” for the func- tion graphed below.
x g(x) g′(x) g′′(x) 0 1 2 3 4
- Sketch functions f for x-values near 1 so that f ( 1 ) = 2 and:
(a) f ′( 1 ) > 0, f ′′( 1 ) > 0 (b) f ′( 1 ) > 0, f ′′( 1 ) < 0 (c) f ′( 1 ) < 0, f ′′( 1 ) > 0 (d) f ′( 1 ) > 0, f ′′( 1 ) = 0, f ′′( 1 −) < 0, f ′′( 1 +) > 0 (e) f ′( 1 ) > 0, f ′′( 1 ) = 0, f ′′( 1 −) > 0, f ′′( 1 +) < 0
- Some people like to think of a concave up graph as one that will “hold water” and of a concave down graph as one which will “spill water.” That description is accurate for a concave down graph, but it can fail for a concave up graph. Sketch the graph of a function that is concave up on an interval, but that will not “hold water.”
4 f ′′^ and the shape of f 255
The function f (x) =
2 π e
− (x 2 −bc 2 ) (^2) defines the Gaussian distribution used extensively in statistics and probability; its graph is a “bell- shaped” curve (see margin).
(a) Show that f has a maximum at x = c. (The value c is called the mean of this distribution.) (b) Show that f has inflection points where x = c + b and x = c − b. (The value b is called the standard deviation of this distribution.)
3. 4 Practice Answers
- See margin figure.
x f (x) f ′(x) f ′′(x) concavity 1 + + − down 2 + − − down 3 − − + up 4 − 0 − down
f ′(x) = 6 x^2 − 30 x + 24 , which is defined for all x. f ′(x) = 0 if x = 1 or x = 4 (critical values). f ′′(x) = 12 x − 30 so f ′′( 1 ) = − 18 < 0 tells us that f is concave down at the critical value x = 1 , so (1, f ( 1 )) = (1, 4) is a relative maximum; and f ′′( 4 ) = 18 > 0 tells us that f is concave up at the critical value x = 4 , so (4, f ( 4 )) = (4, − 23 ) is a relative minimum. A graph of f appears in the margin.
The points labeled (b) and (g) are inflection points.
f ′(x) = 4 x^3 − 36 x^2 + 60 x + 5 ⇒ f ′′(x) = 12 x^2 − 72 x + 60 = 12 (x^2 − 6 x + 5 ) = 12 (x − 1 )(x − 5 ) so the only candidates to be inflection points are x = 1 and x = 5.
- If x < 1 then f ′′(x) = 12 (neg)(neg) > 0
- If 1 < x < 5 then f ′′(x) = 12 (pos)(neg) < 0
- If 5 < x then f ′′(x) = 12 (pos)(pos) > 0
f changes concavity at x = 1 and x = 5 , so x = 1 and x = 5 are both inflection points. A graph of f appears in the margin.
