22.16 - Concavity Test For Finding The Shape of Graph

In the previous section, we we saw the first derivative test for finding local extrema. In this section, we will see the concavity test for finding shape of graph.

First we will see upward concavity. It can be written in 5 steps:
1. The red curve in fig.22.52 below shows the graph of a function f.

Fig.22.52

• We see that, the graph has an upward concavity. In such a situation, we say: f is concave up.

2. Let us analyze three random tangents of f.
Three random points are marked on f. They are: green, magenta and yellow.
• Consider the tangent at the green point.
   ♦ This tangent is drawn in green color.
   ♦ This tangent makes an angle of 110.6^o with the +ve side of the x-axis.
• Consider the tangent at the magenta point.
   ♦ This tangent is drawn in magenta color.
   ♦ This tangent makes an angle of 137.6^o with the +ve side of the x-axis.
• Consider the tangent at the yellow point.
   ♦ This tangent is drawn in yellow color.
   ♦ This tangent makes an angle of 65.2^o with the +ve side of the x-axis.

3. We know that, slope of a line is equal to the tangent of the angle which the line makes with the +ve side of the x-axis.
• So let us write the slopes:
   ♦ Slope of green tangent = tan(110.6) = −2.66
   ♦ Slope of magenta tangent = tan(137.6) = −0.913
   ♦ Slope of yellow tangent = tan(65.2) = 2.16
• We see that:
   ♦ Magenta has a larger slope than green.
   ♦ Yellow has a larger slope than magenta.
• So the slope increases in the order:
Green < Magenta < Yellow

4. Now consider the green, magenta and yellow points.
   ♦ yellow is at the right side of magenta.
   ♦ magenta is at the right side of green.
• So we can write:
For a function which is concave up, the slope of tangent increases as we move from left to right.

5. Recall that, slope of the tangent is given by f '. This f ' is also a function.
• So, if f ' is increasing, the second derivative f '' must be +ve.
• Therefore we can write:
A function f is concave up in an interval (a,b), if f '' is +ve at all points in that interval.

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Now we will see downward concavity. It can be written in 5 steps:
1.The red curve in fig.22.53 below shows the graph of a function f.

Fig.22.53

• We see that, the graph has a downward concavity. In such a situation, we say: f is concave down.

2. Let us analyze three random tangents of f.
Three random points are marked on f. They are: yellow, magenta and green.
• Consider the tangent at the yellow point.
   ♦ This tangent is drawn in yellow color.
   ♦ This tangent makes an angle of 65.2^o with the +ve side of the x-axis.
• Consider the tangent at the magenta point.
   ♦ This tangent is drawn in magenta color.
   ♦ This tangent makes an angle of 137.6^o with the +ve side of the x-axis.
• Consider the tangent at the green point.
   ♦ This tangent is drawn in green color.
   ♦ This tangent makes an angle of 110.6^o with the +ve side of the x-axis.

3. We know that, slope of a line is equal to the tangent of the angle which the line makes with the +ve side of the x-axis.
• So let us write the slopes:
   ♦ Slope of yellow tangent = tan(65.2) = 2.16
   ♦ Slope of magenta tangent = tan(137.6) = −0.913
   ♦ Slope of green tangent = tan(110.6) = −2.66

• We see that:
   ♦ Magenta has a smaller slope than yellow.
   ♦ Green has a smaller slope than magenta.
• So the slope decreases in the order:
Yellow < Magenta < Green

4. Now consider the green, magenta and yellow points.
   ♦ yellow is at the left side of magenta.
   ♦ magenta is at the left side of green.
• So we can write:
For a function which is concave down, the slope of tangent decreases as we move from left to right.

5. Recall that, slope of the tangent is given by f '. This f ' is also a function.
• So, if f ' is decreasing, the second derivative f '' must be −ve.
• Therefore we can write:
A function f is concave down in an interval (a,b), if f '' is −ve at all points in that interval.

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Based on the above two examples, we can write the test for concavity:
Let f be a twice differentiable function in the interval (a,b).
(i) If f '' > 0 for all points in (a,b), then f is concave up.
(ii) If f '' < 0 for all points in (a,b), then f is concave down.

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Now we can learn about inflection point. It can be written in 3 steps:
1. Fig.22.54 below, shows the graph of a function f.

Fig.22.54

• We see that:
   ♦ f is concave up in the interval (a,c)
   ♦ f is concave down in the interval (c,a)
2. Let us analyze the change in sign of f '':
(i) We know that, if f is concave up, all f '' will be +ve.
• So all f '' in (a,c), will be +ve.
(ii) We also know that, if f is concave down, all f '' will be −ve.
• So all f '' in (c,a), will be −ve.
(iii) Such a change in sign for f '' is possible only if:
f '' becomes zero at some point.
(iv) Obviously, that zero point will be at the junction between "concave up portion" and "concave down portion".
• That means, f ''(c) = 0
3. So we can write the definition of inflection point:
Inflection point of a function f is the point at which f changes concavity.
4. Based on the above definition, we can write:
The inflection point in our present case is (c,f(c)).

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Now we will see a solved example
Solved example 22.54
For the function f(x) = x³ − 6x², determine all intervals where f is concave up and all intervals where f is concave down.
Solution:
Step I: Finding the inflection points and dividing the number line
1. First we will write f '(x):
f '(x) = 3x² − 12x

2. Next we will write f ''(x):
f '' = 6x − 12

3. Equating f''(x) to zero, we get:
6x − 12 = 0
⇒ 6x = 12
⇒ x = 2

4. So the only one inflection point is at x = 2

5. Now the number line can be divided into two intervals:
(−∞,2) and (2,∞)

Step II: Finding the sign of f '' in each interval and assessing the concavity
First interval: (−∞,2)
• A convenient number in the first interval is 0.
• f ''(0) = 6(0) − 12 =  −12
• Therefore, the sign of f ''(x) in the first interval is −ve.
• Then f is concave down in the first interval.

Second interval: (2,∞)
• A convenient number in the second interval is 3.
• f ''(3) = 6(3) − 12 = 18 −12 = 6
• Therefore, the sign of f ''(x) in the second interval is +ve.
• Then f is concave up in the second interval.

Step III (optional): Drawing the graph

• The graph is shown in fig.22.55 below:

Fig.22.55

   ♦ f is drawn in red color.
   ♦ f ' is drawn in yellow color
   ♦ f '' is drawn in green color
• The vertical magenta dashed line divides the number line into the two intervals (−∞,2) and (2,∞). The following five facts are in agreement with our discussion on concavity test:
(i) For all inputs from the left of this vertical line, f is concave down.
(ii) For all inputs from the right of this vertical line, f is concave up.
(iii) For all inputs from the left of this vertical line, f ' is decreasing.
(iv) For all inputs from the right of this vertical line, f ' is increasing.
(v) f '' and this vertical line intersects at the x-axis. That means, value of f '' at x = 2 (the point of inflection), is zero.

Solved example 22.55
For the function f(x) = x + sin(2x), x ∈ [−π/2,π/2], determine all intervals where f is concave up and all intervals where f is concave down.
Solution:
Step I: Finding the inflection points and dividing the number line
1. First we will write f '(x):
f '(x) = 1 + 2 cos(2x)

2. Next we will write f ''(x):
f '' = −4 sin(2x)

3. Equating f''(x) to zero, we get:
−4 sin(2x) = 0
⇒ −8 sin x cos x = 0
⇒ sin x = 0 or cos x = 0
• For the equation sin x = 0, the only one solution in [−π/2,π/2], is x = 0
• For the equation cos x = 0, the only two solutions in [−π/2,π/2], are x = −π/2 and x = π/2


4. So the three inflection points are:
x = −π/2, x = 0 and x = π/2
5. Now the given domain [−π/2,π/2] can be divided into two intervals:
[−π/2,0] and [0,π/2]

Step II: Finding the sign of f '' in each interval and assessing the concavity
First interval: [−π/2,0]
• A convenient number in the first interval is −π/8.
• f ''(−π/8) = −4 sin(2(−π/8))
= −4 sin (−π/4) = 4 sin (π/4) = 4/(√2) 
• Therefore, the sign of f ''(x) in the first interval is +ve.
• Then f is concave up in the first interval.

Second interval: [0,π/2]
• A convenient number in the second interval is π/8.
• f ''(π/8) = −4 sin(2(π/8))
= −4 sin (π/4) = −4/(√2) 
• Therefore, the sign of f ''(x) in the second interval is −ve.
• Then f is concave down in the second interval.

Step III (optional): Drawing the graph

• The graph is shown in fig.22.56 below:

Fig.22.56

   ♦ f is drawn in red color.
   ♦ f ' is drawn in yellow color
   ♦ f '' is drawn in green color
• The two vertical magenta dashed lines are the boundaries of the domain [−π/2,π/2]. Those two vertical lines indicate the position of two inflection points x = −π/2 and x = π/2 also. Position of the third inflection point x = 0, is indicated by the y-axis. The following five facts are in agreement with our discussion on concavity test:
(i) For all inputs from between the left vertical line and y-axis, f is concave up.
(ii) For all inputs from between y-axis and the right vertical line, f is concave down.
(iii) For all inputs from between the left vertical line and y-axis, f ' is increasing.
(iv) For all inputs from between y-axis and the right vertical line, f ' is decreasing.
(v) Consider the point of intersection of  f '' with the two vertical lines and the y-axis. All those three points of intersection are at the x-axis. That means, value of f '' at those points (the points of inflection), is zero.

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In the next section, we will see the second derivative test.

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