CHAPTER 2 SUPPLEMENTARY EXERCISES
3.2 FIRST AND SECOND DERIVATIVES
FIRST AND SECOND DERIVATIVES 83 For Exercises 11–15, use the following graph
0 1 2 3 4 5 6 7
–4 –3 –2 –1 0 1 2 3 4
y
x A
B C
D
E
F G
Determine at which of the labeled points the function
11. has a local or endpoint extremum.
12. is increasing.
13. has a point of inflection.
14. is concave down.
15. has an absolute extremum.
y
x Local maximum
Local maximum
Local minimum
Local minimum
First Derivative
Where function values are increasing, the slope and derivative are positive valued. Function values decrease where its slope and derivative are negative. When the derivative is zero, it signals a horizontal tangent. The point at which the derivative vanishes is called astationary pointor acritical point.
↓ Remember that the derivative is the slope, so, naturally, it increases and decreases with the function.
Vanishing derivatives are used to locate extreme points as we will see. However, a van- ishing derivative is not necessarily an extreme point. Functions may have several maxima and minima; the largest (smallest) among them being theabsolute (or global) maximum (minimum) and the otherslocal maxima (minima).
In the graph shown previously, the global maximum is infinite (upward arrow). Likewise, the global minimum is negative infinity (downward arrow).
↓Don’t be baffled by the extensive jargon in use – “extrema,” “critical point,” “local,” “rel- ative,” “global,” “stationary,” and so on. From a calculus perspective, they are usually synonyms and a signal of a horizontal tangent. Exception: they can also refer to end- points, cusps, and such that lack derivatives and horizontal tangents and are considered in advanced texts.
At each value ofxwheref(x) is differentiable, the instantaneous slope of its graph is its derivative there. It is also the slope of the tangent line to the curve at that point.
FIRST AND SECOND DERIVATIVES 85
Example 3.2.1 Slope of a Tangent Line to a Curve
Verify that the slope of y=x2+1 at x=2 is that of the tangent line y=4x−3.
–8 –6 –4 –2 0 2 4 6 8 10 12
–4 –3 –2 –1 0 1 2 3 4
y
x y = 4x – 3
(2, 5) y = x2 + 1
Solution:
The derivative of y=x2+1 is 2x. Therefore, at x=2, the slope of the graph is 4 as is the slope of the tangent line y=4x−3.
The importance, power, and convenience of obtaining slopes of curves, and other prop- erties of functions,y=f(x), simply by differentiation are impressive (when derivatives exist).
In plain language, when a derivative is positive, the function increases asxincreases in the neighborhood ofx. Similarly, a negative derivative means that the function decreases as xincreases. Points at which the derivative vanishes are stationary (critical) points, that is, x0is a stationary (critical) point iff′(x0) =0.
The First Derivative
Values ofxfor whichf′(x)>0 correspond tof(x) increasing.
Values ofxfor whichf′(x)<0 correspond tof(x) decreasing.
Values ofxfor whichf′(x) =0 correspond tof(x) being stationary.
A first derivative that vanishes identifies a critical point as it signals a horizontal tangent (slope is zero). A horizontal tangent can signal a maximum or minimum. Sometimes, it signals an inflection point.
These following diagrams are illustrative!
y
x
Maximum at the horizontal tangency
y
x
Minimum at the horizontal tangency
y
x
Inflection point at the horizontal tangency
y
x
Inflection point at the horizontal tangency
Note that while the slope off(x) vanishes and its sign changes on passing through a max- imum or minimum, the sign of the slope does not change on passing through an inflection point.
It is a necessity for the slope to change sign on passing through a maximum or minimum.
Therefore, one cannot always depend solely on a vanishing derivative to signal a maximum or minimum. An example illustrates:
Example 3.2.2 Local Extrema
Determine all local extrema of f(x) =2x3−3x2+5 Solution:
Firstly, determine the critical points by setting the first derivative to zero.
f′(x) =6x2−6x=6x(x−1) =0
The critical points, where extrema are possible, are at x=0 and x=1. The table orga- nizes the signs of the factors and their product.
6x – +
– –
–
+
x – 1 +
f′(x) = 6x(x − 1) + +
0 1
x < 0 0 < x < 1 x > 1
FIRST AND SECOND DERIVATIVES 87 The sign change of f′(x)(positive to negative) about x=0 indicates a maximum. The change in sign of f′(x)(from negative to positive) about x=1 indicates a minimum.
Therefore, f(0) =5 is a local maximum and f(1) =4 a local minimum.
Note that while the derivative elegantly signals a local extremum, it is silent as to whether it may also be an absolute maximum or minimum. For a maximum or minimum of a func- tion,f(x), to occur at a critical pointx=a, it is anecessary conditionthat its derivative vanishes there, that is, thatf′(a) =0. However, it is not asufficient conditionfor extrema as illustrated in the following example.
Example 3.2.3 More on Local Extrema
Show that f(x) =x3lacks extrema.
Solution:
For possible extrema, set f′(x) =0. Here f′(x) =3x2. Therefore, the only critical point is at x=0.
Next, investigate the sign of the derivative about x=0. The derivative 3x2 is positive (increasing function) both when x<0 and when x>0. Therefore, without a sign change at x=0, there is neither a maximum nor a minimum. (Soon we will learn that it is an inflection point.)
Second Derivative
The derivative of a function, sayy=f(x),yields another function we have denoted byf′(x).
Now,f′(x)as a function can have a derivative – called a second derivative off(x),f′′(x). The second derivative also aids in determining the behavior of functions. The second derivative, the “derivative of the first derivative,” is therate of changeof the first derivative. Graphi- cally, it is the rate of change of the slope of the tangent line. Unfortunately, it is not easy to depict in a sketch.
The slopes of the tangents, being first derivatives, their rate of change is the second derivative (the rate of change of the first derivative). Consider the local maximum, depicted in the concave down curve that follows, “it spills water.” Observe the change in slopes of tangents as they pass through the maximum. Clearly, their slopes continuously decrease as xincreases. Here, in a maximum, the rate of change of the slope of the tangent is negative.
y
x
Maximum
Second derivative is negative
y
x Minimum
Second derivative is positive
Similarly, for a local minimum (concave up curve on the top right), the changes in slopes of tangents increase asxincreases. The second derivative is positive for a minimum, “it holds water.”
↓ Note the decreasing slope of the tangent (left figure) from positive values to zero and then to negative values. Similarly, note the increasing slope of the tangent, from nega- tive values to zero to positive values (right figure). It is correct to associate the second derivative with the change in slope of the tangent line.
Concavity
Values ofxfor whichf′′(x)>0 correspond tof(x) concave up (“holds water”).
Values ofxfor whichf′′(x)<0 correspond tof(x) concave down (“spills water”).
Inflection Points
Sometimes, a second derivative vanishes at a point. This means its sign has changed from positive to negative (or vice versa) asxincreases through the vanishing point. For the second derivative to change in sign from positive to negative (or vice versa), it must pass through zero. It is a signal for aninflection pointas in the following example.
Example 3.2.4 Points of Inflection
Show that y=x3+1 (sketch in Section 3.1) has a horizontal tangent at its inflection point at (0, 1).
Solution:
The first and second derivatives are f′(x) =3x2and f′′(x) =6x, respectively. Both deriva- tives vanish at x=0. However, as the first derivative does not change sign (positive for all x here), there can neither be a maximum nor a minimum at x=0.
Clearly, for x<0, the second derivative is negative and for x>0, it is positive. The sign change in the second derivative on passing through x=0 signals a change in concavity. It corresponds to an inflection point at (0, 1).
While all inflection points have tangents, they are not always horizontal. Consider y=x3−4x. Here,y′(x) =3x2−4 andy′′(x) =6x. The second derivative vanishes atx=0, and asy′′(x)changes sign asxchanges from negative to positive values, there is an inflec- tion point atx=0. Now, atx=0,yis also zero, so the curve passes through the origin. Its slope at the originy′(0) = −4; which is also the slope of the tangent there as shown in the figure.
FIRST AND SECOND DERIVATIVES 89
–20 –15 –10 –5 0 5 10 15 20
–4 –3 –2 –1 0 1 2 3 4
y
x Tangent line
y = –4x
y = x3 – 4x
The maxima and minima in the preceding examples were easy to identify from their graphs, if not from the functions themselves. However, in practical situations, graphs may be neither available nor convenient. The question arises as to whether one can identify a horizontal tangent with certainty as a maximum, a minimum, or an inflection point. Remark- ably, the answer is yes! And a (sufficient) test is simple!
The Second Derivative and Local Extrema
Letf(x) be differentiable on an open interval containingx=aand f′(a) =0.
Iff′′(a)>0,f(x) has a local minimum atx=a Iff′′(a)<0,f(x) has a local maximum atx=a.
(Iff′′(a) =0, the test fails and the changing slope of the first derivative about x=adetermines the character of the function there).
Example 3.2.5 The Second Derivative and Local Extrema
Locate and identify maxima and minima of y=x3−6x2+9x using its derivatives. (A graph appears early in the previous section.)
Solution:
In this case, y′=3x2–12x+9. Setting y′to zero (and factoring) yields
y′=3(x−1)(x−3) =0. Hence, the critical points (corresponding to horizontal tangents) occur at x=1 and x=3.
The second derivative is y′′=6x−12. At x=1,y′′<0. This negative second derivative indicates a maximum as the curve is “concave down” about x=1.
At x=3, the second derivative is positive. This indicates a minimum as the curve is
“concave up” about x=3.
Therefore, there is a local maximum at (1, 4) and a local minimum at (3, 0).
Example 3.2.6 Quadratics and Local Extrema
Locate and identify extrema of the quadratic y=ax2+bx+c. (a≠0) Solution:
In this case, y′=2ax+b. Setting y′to zero yields x= −b
2a as a possible extremum.
The second derivative is y′′=2a. Therefore, when a<0, the function is concave down at x= −b
2a and the function is a maximum there. When a>0, the function is concave up at x= −b
2a and the function is a minimum there. You may recall this as the axis of symmetry and parabola opening according to the sign of a.
EXERCISES 3.2
In Exercises 1–4, identify graphs that depict 1. a positive first derivative for allx.
2. a negative first derivative for allx.
3. a positive second derivative for allx.
4. a negative second derivative for allx.
(a) (b) (c) (d)
(e) (f) (g) (h)
In Exercises 5–8, sketch the described function.
5. The graph is concave down for allx, (2, 5) is on the graph,f(0) =1 andf′(2) =0.
6. f′′(x)<0 for allx,f(0) = −1,f′(1) =0, and (1, 0) on the graph.
FIRST AND SECOND DERIVATIVES 91 7. f(1) =4∕3,f(3) =0,f′(x)>0 on(−∞,1) ∪ (3,∞), andf′(x)<0 on (1, 3).
8. f′(x)>0 on(−∞,0) ∪ (2,∞),f′(x)<0 on (0, 2).(2,−2)and (0, 2) are on the graph.
9. Using the following sketch, enter+,−, or 0 in the table as appropriate. Assume a local minimum at B and inflection point at C.
f f′ f″
–10 –5 0 5 10 15 20
–3 –2 –1 0 1 2 3
y
x A
B
D C
A B C D
In Exercises 10−15, use the following graph
–10 –8 –6 –4 –2 0 2 4 6 8 10
–2 –1 0 1 2 3 4 5
y
x y = f′(x)
(2, –6)
10. Why mustf(x) be decreasing atx=2?
11. Why mustf(x) be increasing atx=5?
12. Why mustf(x) have an inflection point atx=0?
13. Where doesf(x) have a local maximum?
14. Determine the two values ofxwheref(x) has local minima.
15. Iff(2) = −2∕3 determine the equation of the tangent to the curve atx=2.