Concavity And Inflection

1.  Concavity

Let f be differentiable on (a, b). See Fig. 1. The graph of f is bending upward. We say that the graph of f is concave up. We note that as x increases from the near right of a to the near left of b, the tangent line to the graph of f at x turns counterclockwise, which means that its slope increases. Now,
the slope of the tangent line at x is f '(x). So, as x increases in (a, b), f '(x) increases.

Let g be differentiable on (a, b). See Fig. 2. The graph of g is bending downward. We say that the graph of g is concave down. We note that as x increases from the near right of a to the near left of b, the tangent line to the graph of g at x turns clockwise, which means that its slope decreases. So, as x
increases in (a, b), g'(x) decreases.

Definition 1
Suppose the function f is differentiable on (a, b). We say that (the graph of ) f is concave up on (a, b) if f '(x) is increasing there. Similarly, we say that (the graph of ) f is concave down on (a, b) if f '(x) is decreasing there.



Remarks 1

i.    Concavity is defined only for differentiable functions.

ii.   It can be shown that if a function is concave up on an open interval, then any tangent line to its graph on that interval lies below the graph, and any chord line on that interval lies above the graph.
     

iii.  Analogously, if a function is concave down on an open interval, then any tangent line to its graph on that interval lies above the graph, and any chord line on that interval lies below the graph.


2.  Inflection Points

The graph of y = f(x) = x³ is shown in Fig. 3. We have f '(x) = 3x² and thus f ''(x) = 6x. Clearly, f ''(x) < 0 for all x < 0 and f ''(x) > 0 for all x > 0. Remember, f '' is the (first) derivative of f '. So, by the first-derivative test, f ' is decreasing on (– ¥, 0) and increasing on (0, ¥). Consequently, by definition, f is concave down on (– ¥, 0) and concave up on (0, ¥). This means that f changes concavity at x = 0. We say that f changes concavity  at a point x₁ if its concavity is opposite on opposite sides of x₁.

The graph of y = g(x) = x^1/3 is shown in Fig. 4. We have g'(x) = (1/3)x^–2/3 and thus g''(x) = –2/(9x^5/3). Clearly, g''(x) > 0 for all x < 0 and g''(x) < 0 for all x > 0. So, g' is increasing on (­– ¥, 0) and decreasing on (0, ¥). Consequently, g is concave up on (– ¥, 0) and concave down on (0, ¥). This means that g changes concavity at x = 0.

The graph of h is shown in Fig. 5. We see that h changes concavity at x = 0.

Now, we want to call the point where the function changes concavity an inflection point, like the point x = 0 for f and g, but unlike the point x = 0 for h because h has a sharp point there. We remark that:

i.  All three functions change concavity at x = 0.

ii.  We have f '(0) = f '(x)|_x=0 = 3x²|_x=0 = 0, ie, f is differentiable at x = 0. Thus, the tangent line to f at
x = 0 has equation ( y – 0) = 0(x ­– 0), or y = 0, which is the x-axis. As for g, since g'(x) =(1/3)x^–2/3 = 1/(3x^2/3), g isn't differentiable at x = 0. However, the y-axis is a tangent line to g at x = 0. We see that g has a vertical tangent line at x = 0. As for h, it has a sharp point at x = 0, hence it isn't differentiable and doesn't have a tangent line at x = 0.

Of course we'll include the change of concavity as a requirement in the definition of an inflection point.
If we include differentiability, we'll eliminate h but we'll also eliminate g. We see that the inclusion of the existence of a tangent line is appropriate. Therefore, we'll define an inflection point  as a point where the function has a tangent line and changes concavity.


Definition 2

A point x₁ is called an inflection point of a function f if f has a tangent line at x₁ and f changes
concavity at x₁.


Remarks 2

Suppose x₁ is an inflection point of f.

i.  f either is differentiable at x₁ or has a vertical tangent line there; this is clear from the definition.

ii.  The tangent line to the graph of f at x₁ crosses the graph of f at that point. If the tangent is vertical, this property is obvious. If the tangent is non-vertical, this property means that the tangent is above the graph on one side of x₁ and below it on the other side. 


3.  How Concavity And Inflection Points Relate To The
     Second Derivative

Refer to Fig. 3. For x > 0, f ''(x) = 6x > 0 and f is concave up. For x < 0, f ''(x) < 0 and f is concave down.

Now, suppose x₁ is an inflection point of f. As seen in Part 2 above, x₁ = 0 is an inflection point of f(x) = x³ and also of g(x) = x^1/3; we see that f ''(x) = 6x exists at x₁ = 0 while g''(x) = –2/(9x^5/3) doesn't exist at x₁ = 0. We observe that f ''(x₁) = f ''(0) = 0.


Theorem 1

i.   Let f be twice differentiable on (a, b).
     If f ''(x) > 0 on (a, b), then f is concave up on (a, b)
     If f ''(x) < 0 on (a, b), then f is concave down on (a, b).

ii.  If x₁ is an inflection point of f and f ''(x₁) exists, then f ''(x₁) = 0.


Proof
i.  If f ''(x) > 0 on (a, b), then f ' is increasing there, so f is concave up on there. If f ''(x) < 0 on (a,b), then f ' is decreasing there, so f is concave down on there.

ii.  Suppose x₁ is an inflection point of f and f ''(x₁) exists. Since f changes concavity at x₁, f ' changes from increasing to decreasing or vice versa as x passes thru x₁. As f ' is differentiable at x₁, it's continuous there. Thus, f '(x₁) is a local maximum or a local minimum of f '. Consequently, by
 f ''(x₁) = 0.

4.  Determining Concavity And Finding Inflection Points

Part i of the above theorem gives us a straightforward way to determine the intervals of concavity of
a twice differentiable  function f.

Part ii of the theorem provides us this insight: inflection points of a twice differentiable  function f can occur only at points x where f ''(x) = 0. (If f ''(x) ¹ 0, then x can't be an inflection point, because if it were then we would have f ''(x) = 0.) That's similar to the situation where a local extremum can occur only at points x where f '(x) = 0 or f '(x) doesn't exist (critical point) or x is an endpoint; . There, we see that not every such point yields a local extremum. Here, the situation is similar: not every point x where f ''(x) = 0 is an inflection point. For example, for f(x) = x³ and x₁ = 0, as shown in Fig. 3, we have f ''(x₁) = 0 and x₁ is an inflection point of f; however, for g(x) = x⁴ and x₁ = 0, as shown in Fig. 6, we have g'(x) = 4x³ and g''(x) = 12x², so g''(x₁) = 0, but x₁ isn't an inflection point of g, because g doesn't change concavity at x₁ ( g is concave up on both sides of x₁ and
as a matter of fact on every interval).

So, to find inflection points of a twice differentiable  function f, first we identify all points x where f ''(x)= 0, then we check to see at which of those points f changes concavity. The determining of the intervals of concavity and the finding of the inflection points of a twice differentiable  function is
illustrated in the following example.


Example 1

Let f(x) = x⁴ – 4x².

b.  Locate and classify the local extrema of f, using a chart.
c.  Determine the intervals of concavity and find the inflection points of f, using a chart.
d.  Combine the informations in part  b, and c in a chart. Sketch the graph of f.

Solution

 

   

d.

    The graph of f is sketched in Fig. 10.

EOS


Chart Showing Concavity And Inflection Points




Chart Summarizing The Behavior Of A Function

The chart for f in Fig. 9 summarizes the behavior of f: intervals of increase and decrease, local extrema, intervals of concavity, and inflection points. All the critical points and all the points x where f ''(x) = 0 are placed in the row for x in increasing order. The meanings of the arcs in the row for f(x)
in the charts in Figs. 8 and 9 are shown in the table in Fig. 11.

5.  Distinction Between Inflection Points, Critical Points,   And Local Extrema

Let f be a twice differentiable function. The following distinction should be clear.

An inflection point of f is a point x of dom ( f ) where f changes concavity. It can be only at points x where f ''(x) = 0.

A critical point of f is a point x of dom ( f ) where f '(x) = 0 (non-existence of f '(x) is excluded because f is twice differentiable).

A local extremum is a value of f that is the maximum or minimum of all the values f(x) of f for x on an interval of dom ( f ). So it's a point in range ( f ). It can occur only at critical points of f, i.e., at points x where f '(x) = 0, and endpoints of dom ( f ).

For the function f(x) = x³ in Fig. 3, the point x = 0 is an inflection point. As f '(0) = f '(x)|_x=0 = 3x²|_x=0 =
0, the point x = 0 is also a critical point. For the function f(x) = (1/30)x⁶ – (1/3)x⁴ as summarized in Fig. 9, the inflection points are x = –2 and x = 2, but f '(–2) ¹ 0 and f '(2) ¹ 0; thus x = 2 and x = –2
aren't critical points. We've just seen that an inflection point may or may not coincide with a critical
point.



Problems & Solutions


1.  Let:

 

Solution



    



    

    Thus, f is concave up on (– ¥, –2) and (2, ¥) and concave down on (–2, 0) and (0, 2), and the
    inflection points of f are x = –2 and x = 2; x = 0 isn't an inflection point.

d.

    

    

2.  Sketch the graph of a twice differentiable function f where f ''(x) < 0 for x < 2, f(2) = – 1, f '(2) =
     1, f ''(2) = 0, f ''(x) > 0 for x > 2, and f(4) = 3.

Solution



The equation of the tangent line to the graph at x = 2 is y = (–1) + (1)(x – 2), or y = x – 3. The graph
is below the tangent line to the left of x = 2, and above it to the right of x = 2.







3.  Prove, using the mean-value theorem, that for a twice differentiable function f, one can't have
     f(2) = –1, f '(2) = 1, f ''(x) > 0 for all x > 2, and f(4) = 1.

Note

See the graph in the previous Problem & Solution.

Solution

Since f ''(x) > 0 on (2, ¥), f ' is increasing on (2, ¥); so f '(x) > f '(2) for all x > 2. If f(4) = 1, then,
because f is differentiable on [2, 4], we would have, by the mean-value theorem, ( f(4) – f(2))/(4 – 2)
= f '(c) for some c where  2 < c < 4. So f '(c) = (1 – (–1))/2 = 1 = f '(2), which is a contradiction.




4.  Prove that if a function is concave up on an open interval, then any tangent line to its graph on tha interval lies below the graph.

Note

The graphs below will help in the proof.



Solution

Let f be a concave-up function on (a, b) and x₁ where a < x₁ < b be arbitrary. The equation of the tangent line to the graph of f at x₁ is y = t(x) = f(x₁) + f '(x₁)(x – x₁). Let g(x) = f(x) – t(x) = f(x) – ( f(x₁) + f '(x₁)(x – x₁)) on (a, b). Note that g(x) = f(x) – f '(x₁)x + C, where C = x₁ f '(x₁) – f(x₁) is a constant, and that g(x₁) = 0.

We have g'(x) = f '(x) – f '(x₁). So g'(x₁) = 0. Because f is concave up on (a, b), f ' is, by definition, increasing there. Thus, g'(x) < 0 for all x where a < x < x₁, and g'(x) > 0 for all x where x₁ < x < b.
Hence, by the first-derivative test, g has a minimum value at x₁, which is g(x₁) = f(x₁) – ( f(x₁) +
f '(x₁)(x₁ – x₁)) = 0, on (a, b). We've got g(x) ³ g(x₁) = 0 for all x in (a, b).

Now, suppose a < x < x₁. If g(x) = g(x₁), then, since g is continuous on [x, x₁] and differentiable on (x, x₁), Rolle's theorem would imply that there exists c where x < c < x₁ such that g'(c) = 0, which contradicts the fact that it must be that g'(c) < 0. So we must have g(x) > g(x₁). Similarly, we have
g(x) > g(x₁) if x₁ < x < b.

Therefore, g(x) > 0 on (a, x₁) U (x₁, b) and g(x₁) = 0, or f(x) > t(x) on (a, x₁) U (x₁, b) and f(x₁) = t(x₁), which mean that the tangent line to the graph of f at x₁ lies below the graph. Since x₁ in (a, b) is arbitrary, the proof is complete.