5.1 Curve Sketching
5.1.3 Convexity
136 Chapter 5. Applications of Differentiation
Solving f0(x)=0, we get the critical numbers of f: x1=0 andx2=3.
From the table, we see that
• the critical numberx1=0 is not a local extremizer of f;
• the critical numberx2=3 is a local minimizer of f.
5.1. Curve Sketching 137
bending
up bending
down
Figure 5.10(a) Figure 5.10(b)
Alternatively, if the curve is the graph ofy= f(x) where f is a differentiable function, the graph is bending up (respectively bending down) means that the graph is always above (respectively always below) the tangent lines.
Remark In many books, instead of bending up and bending down, the termsconcave upandconcave down respectively are used.
Bending up and bending down are properties of curves. Below are properties of functions corresponding to these geometric properties.
Definition Let f be a function that is defined and differentiable on an open interval (a,b). We say that
• f isstrictly convexon (a,b) if f0is increasing on (a,b);
• f isstrictly concaveon (a,b) if f0is decreasing on (a,b).
Since f0 is the slope function, f is strictly convex on (a,b) means that the slope is increasing and so in the interval (a,b), the graph of f is bending up. Similarly, f is strictly concave means that in (a,b), its graph is bending down.
Terminology For simplicity, instead of saying “strictly convex”, we will say “convex” etc.
The next theorem describes a simple way to find where a function is convex or concave. The method is to consider the sign of f00.
Theorem 5.1.4 Let f be a function that is defined and is twice differentiable on an open interval (a,b).
(1) If f00(x)>0for allx∈(a,b), then f is convex on (a,b).
(2) If f00(x)<0for allx∈(a,b), then f is concave on (a,b).
Proof We give the proof for (1). The proof of (2) is similar to that for (1).
Note that f00is the derivative of f0. If f00(x)>0 for allx∈I, that is, (f0)0(x)>0 for allx∈(a,b), then by Theorem 5.1.1, f0is increasing on (a,b), that is, f is convex on (a,b).
Example Let f :R−→Rbe the function given by
f(x)=27x−x3. Find the interval(s) on which f is convex or concave.
138 Chapter 5. Applications of Differentiation
Explanation
• The question is to find maximal open interval(s), if any, on which f is convex or concave.
• The given function f is a “nice” function (a polynomial function). It can be differentiated any number of times: f(n)(x) exists for all positive integers n and for all real numbers x. In particular, f is twice differentiable onR. To apply Theorem 5.1.4, we have to solve inequalities f00(x) > 0 and f00(x) < 0.
This is done by setting up a table.
Solution Differentiating f(x), we get f0(x) = d
dx(27x−x3)
= 27−3x2. Differentiating f0(x), we get f00(x) = d
dx(27−3x2)
= −6x. (−∞,0) (0,∞)
−6 − −
x − +
f00 + −
• On the interval (−∞,0), f is convex.
• On the interval (0,∞), f is concave.
Remark
• When we consider “a function is convex/concave on an interval”, unlike increasing/decreasing, we do not include the endpoint(s) of the interval. This is because the concept is defined for open intervals only.
Note that for a function f whose domain is a closed and bounded interval [a,b], f0(x) is undefined when x=aorb.
• There is a more general definition for convex/concave functions. The definition does not involve f0 and it can be applied to closed intervals also.
Example Let f :R−→Rbe the function given by
f(x)= x4−4x3+5.
Find where the graph of f is bending up or bending down.
ExplanationThis question is similar to the last one. The graph of f is bending up (or down) means that f is convex (or concave). So we have to find intervals on which f00is positive (or negative).
Solution Differentiating f(x), we get f0(x) = d
dx(x4−4x3+5)
= 4x3−12x2. Differentiating f0(x), we get f00(x) = d
dx(4x3−12x2)
= 12x2−24x
= 12x(x−2).
(−∞,0) (0,2) (2,∞)
12x − + +
x−2 − − +
f00 + − +
• On the intervals (−∞,0) and (2,∞), the graph of f is bending up.
5.1. Curve Sketching 139
• On the interval (0,2), the graph of f is bending down.
Definition Let f be a function and let x0 be a real number such that f is continuous atx0 and differentiable on both sides of x0. If f is convex on one side of x0 and concave on the other side, then we say that x0 is an inflection numberof f.
Explanation
• The condition “f is differentiable on both sides of x0” means that there is an open interval in the form (a,x0) and an open interval in the form (x0,b) such that f0(x) exists for allx∈(a,x0)∪(x0,b).
• The condition “f is convex on one side of x0and concave on the other side” means that there is an open interval in the form (α,x0) and an open interval in the form (x0, β) on which f is convex on one of them and concave on the other, that is, there is a change of convexity atx0.
Suppose that x0is an inflection number of a function f. By definition, on one side of the point x0,f(x0) , the graph of f is bending up and on the other side, the graph is bending down. That is, there is a change of bending at the point x0,f(x0)
.
Terminology Suppose thatx0 is an inflection number of a function f. Then the point x0,f(x0)
is called an inflection pointof the graph of f.
In the following example, the function f is discussed in a previous example. Below we just copy part of the table obtained in the solution there.
Example Consider the function f :R−→Rgiven by
f(x)=x4−4x3+5.
From the table
(−∞,0) (0,2) (2,∞)
f00 + − +
we see that the inflection numbers of f are 0 and 2.
RemarkWe can also say that the inflection points of the graph are (0,5) and (2,−11).
The next result gives a necessary condition for inflection number.
Theorem 5.1.5 Let fbe a function and letx0be a real number such that f is differentiable on an open interval containingx0and that f00(x0)exists. Suppose thatx0is an inflection number of f. Then we have f00(x0)=0.
Proof By symmetry, we may assume that f is convex on the left-side ofx0and concave on the right-side ofx0, that is, there exist real numbersaandbwitha< x0<bsuch that f0is increasing on (a,x0) and decreasing on (x0,b), which by continuity of f0atx0, implies that
f0(x)< f0(x0) for allx∈(a,x0)∪(x0,b).
140 Chapter 5. Applications of Differentiation
Thus, the function f0has a local maximum atx0. Hence by Theorem 5.1.3 (and using the assumption that the derivative of f0atx0exists), the derivative of f0atx0is 0, that is, f00(x0)=0.
RemarkThe converse of Theorem 5.1.5 is not true: if f00(x0)=0,x0may not be an inflection number of f. Example Let f :R−→Rbe the function given by
f(x)= x4. Then we have f0(x) = 4x3
f00(x) = 12x2 (−∞,0) (0,∞)
f00 + +
Although f00(0) = 0, the number 0 is not an inflection number of f. This is because f is convex on (−∞,0) as well as on (0,∞).
RemarkThe function f is convex on (−∞,∞). Figure 5.11
Terminology
• If f0(x0) = 0, we say that x0 is a stationary number of f. However, if f00(x0) = 0, we do not have a specific name forx0.
• For local extremizers, there are two types: local maximizers and local minimizers. Correspondingly, there are also two types of inflection numbers. However, we do not have specific names to distinguish the two types.
Example Let f :R−→Rbe the function given by
f(x)= x4−6x2+5x−6.
Find the inflection point(s) of the graph of f.
ExplanationTo find the inflection points of the graph, first we find the inflection numbers of the function. For that, we solve the equation f00(x)=0. By Theorem 5.1.5, solutions to this equation include all the possible can- didates for inflection numbers. However, for each of these candidates, we have to check whether the convexity of f are different on the left-side and right-side of it.
Solution Differentiating f(x), we get f0(x) = d
dx(x4−6x2+5x−6)
= 4x3−12x+5.
Differentiating f0(x), we get f00(x) = d
dx(4x3−12x+5)
= 12x2−12
= 12(x+1)(x−1).
Solving f00(x)=0, we get two solutions: x1 =1 andx2=−1.
5.1. Curve Sketching 141
(−∞,−1) (−1,1) (1,∞)
12 + + +
x+1 − + +
x−1 − − +
f00 + − +
From the table, we see that f is convex on (−∞,−1), concave on (−1,1) and convex on (1,∞). Hencex1 = 1 andx2=−1 are the inflection numbers of f.
The inflection points of the graph are 1,f(1)
=(1,−6) and −1, f(−1)
=(−1,−16).
To determine the nature of critical numbers, we can use the First Derivative Test discussed in the last subsection. Below, we discuss an alternative way using second derivatives.
Second Derivative Test Let fbe a function and letx0be a real number such that fis differentiable on an open interval containingx0. Suppose thatx0is a critical number of f, that is, f0(x0)=0.
(1) If f00(x0)<0, thenx0is a local maximizer of f (in fact, we have f(x0) > f(x) for allxsufficiently close to and different fromx0).
(2) If f00(x0)>0, thenx0is a local minimizer of f (in fact, we have f(x0) < f(x) for all xsufficiently close to and different fromx0).
ExplanationBelow we give a proof for (1). To prove (2), we can use the method for (1). Alternatively, we can apply (1) to the function−f because (−f)00(x0)<0 in this case.
Proof It suffices to prove (1). Suppose that f00(x0) < 0. We want to show that f is increasing on the left-side ofx0and decreasing on the right-side.
By definition, together with the condition f0(x0)=0, we have 0> f00(x0)= lim
h→0
f0(x0+h)
h ,
which implies that f0(x0+h)>0 ifhis sufficiently close to and less than 0, that is, f0(x)>0 ifx=x0+his sufficiently close to and less thanx0.
Hence, by Theorem 5.1.1, f is increasing on the left-side of x0. Similarly, f is decreasing on the right-side of x0. Therefore, by the continuity of f atx0, we see that f(x0) > f(x) for allxsufficiently close to and different
fromx0.
Remark
• To determine the nature of a critical number using the Second Derivative Test, we consider the sign of f00at the critical number. If we apply the First Derivative Test, we consider the sign of f0on the left-side and the right-side of the critical number.
• If f00(x0) = 0, we can’t apply the Second Derivative Test. At x0, the function f may have a local maximum, a local minimum or neither. See the last example in this subsection.
142 Chapter 5. Applications of Differentiation
• Some students have the following conjecture:
Suppose f0(x0)=0and x0is not a local extremizer of f , then x0is an inflection number of f .
For “nice” functions (for example, polynomial functions), the conjecture is correct. However, we can construct weird functions with “weird” critical point (see Figure 5.7).
However, when we consider nature of a critical number, there is no need to discuss whether it is an inflection number because critical numbers are related to first derivatives whereas inflection numbers are related to second derivatives.
Below, we redo a previous example using the Second Derivative Test.
Example Let f :R−→Rbe the function given by
f(x)=27x−x3. Find and determine the nature of the critical number(s) of f. Solution Differentiating f(x), we get f0(x) = d
dx(27x−x3)
= 27−3x2
= 3(3+x)(3−x).
Solving f0(x)=0, we get the critical numbers of f: x1=−3 andx2=3.
Differentiating f0(x), we get f00(x) = d
dx(27−3x2)
= −6x
• At x1=−3, we have f00(−3)=18>0; therefore,x1is a local minimizer of f.
• At x2=3, we have f00(3)=−18<0; therefore,x2is a local maximizer of f.
Example Let f,gandhbe functions fromRtoRgiven by
f(x)= x4, g(x)=−x4, h(x)= x3.
It is clear thatx1=0 is a critical number of f,gandh. Moreover, we have f00(0)=g00(0)=h00(0). However,
• atx1 =0, f has a local minimum;
• atx1 =0,ghas a local maximum;
• atx1 =0,hdoes not have a local extremum.
y=x4 y=−x4 y= x3
Figure 5.12(a) Figure 5.12(b) Figure 5.12(c)
5.1. Curve Sketching 143