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1800-102-2727We can figure out whether the function is increasing or decreasing with the use of derivatives. Now another thing that comes to our mind is the shape of the curve which means whether the curve is facing upward or downward. The upward and downward tendency of the curve can be given by means of the concavity of the curve. So let us discuss the concavity of the curve and then we will extend the concept of concavity to understand the Point of Inflection of the curve.
Table of Contents
In the figures shown above both the functions are in such a way that, as the value of x increases the slope of the tangent given by its derivative drawn also increases. In the first case, with an increase in x slope changes from a less positive value to a more positive value hence increasing and in the second case with an increase in x the slope changes from a more negative value to a less negative value hence again the slope is increasing. Hence we can say that the derivative of both functions is an increasing function. Thus, whenever the derivative of a function is an increasing function then the function will have its shape as concave up.
Similarly, In the figures shown here both the functions are in such a way that as the value of x increases the slope of the tangent drawn decreases. In the first case slope with an increase in x slope changes from a more positive value to a less positive value hence decreasing and in the second case with an increase in x the slope changes from a less negative value to more negative value hence again the slope is decreasing. Hence we can say that the derivative of both functions is a decreasing function. Thus, whenever the derivative of a function is a decreasing function then the function will have its shape as concave down.
Now what is the importance of learning the Test for Concavity of a function? You can easily understand it here. What to do if we just want to check the concavity of a given function without drawing its graph? Here comes the use of the Test for Concavity of function. We should determine the first and second derivatives of the given function to apply the Test for Concavity.
By the definition, we concluded that for a function f to be concave up it derivative i.e. f' must be increasing, and for f' to be increasing derivative of f' i.e. f'' must be positive. Thus if there exists a function f such that it is twice differentiable and f''>0 then the given function will be concave up.
If f''>0 for the given interval I, then f is concave up.
Similarly, if there exists a function f such that it is twice differentiable and f''<0 then the given function will be concave down.
If f''<0 for the given interval I, then f is concave down.
Finding out the concavity of a given function with the above-explained logic is known as the Test for Concavity.
Sign of f' |
Nature of f |
Sign of f'' |
Concavity of f |
Positive |
Increasing |
Positive |
Concave up |
Positive |
Increasing |
Negative |
Concave down |
Negative |
Decreasing |
Positive |
Concave up |
Negative |
Decreasing |
Negative |
Concave down |
Point of inflection comes into consideration when a function switches its concavity in the given interval. Hence in a given interval for a segment if f''>0 and for another segment if f''<0, then the function will be changing its concavity and the point at which this phenomenon occurs will be called the Point of Inflection of f. To check the point of inflection we look at the points where the f''=0 or f'' is undefined. But it should also be kept in mind that when f''=0 or undefined it is not always necessary that the function will be changing its concavity.
Hence summarily when the function f is continuous at a point a and also changes its concavity at a, then (a,f(a)) will be the Inflection Point of the function f.
Therefore, the point (a,f(a)) is an inflection point.
Example 1: Which are the points of inflection on the shown curve?
Solution: Let us analyze the curve for different segments.
$0-2\Rightarrow concavedown$
$2-4\Rightarrow concaveup$
$4-6\Rightarrow concavedown$
$6-7\Rightarrow concaveup$
$7-9\Rightarrow concavedown$
Thus, the points where the concavity is changing are 2, 4, 6 and 7. Hence these points will be the inflection points.
Example 2: Analyze the given exponential function f(x)=kex and comment on its existence of inflection points.
Solution:
Given $f\left(x\right)=k{e}^{x}$
Finding the first derivative of the given function,
$f\text{'}\left(x\right)=k{e}^{x}$
Now finding the second derivative of the given function,
$f\text{'}\text{'}\left(x\right)=k{e}^{x}$
Now checking for the sign of f''(x)
ex is always a positive function for any value of x in its domain hence when k is positive f''(x) will always be positive and when k is negative f"(x) will be negative. Thus for any fixed value of k the function f''(x) will not be changing its sign. Therefore f(x) do not have any points of inflection for any value of k.
Example 3: find the intervals where the function $f\left(x\right)={x}^{4}-6{x}^{3}-108{x}^{2}+57x+2$ is concave up and concave down. Also, find Points of Inflection if they exist.
Solution:
Given $f\left(x\right)={x}^{4}-6{x}^{3}-108{x}^{2}+57x+2$
Finding the first derivative of the given function,
$f\text{'}\left(x\right)=4{x}^{3}-18{x}^{2}-216{x}^{+}57$
Now finding the second derivative of the given function,
$f\text{'}\text{'}\left(x\right)=12{x}^{2}-36{x}^{-}216$
$f\text{'}\text{'}\left(x\right)=12({x}^{2}-3{x}^{-}18)$
$f\text{'}\text{'}\left(x\right)=12(x+3)(x-6)$
Now checking for the sign of f''(x)
For x (-3,6), f''(x)<0 so, the function will be concave downwards
For x (-∞,-3)(6,∞), f''(x)>0 so, the function will be concave upwards
Hence at x=-3 and x=6 , f''(x) changes sign so these will be the inflection points.
Example 4: Is x=0 an inflection Point for the function $g\left(x\right)={x}^{\frac{1}{3}}.$
Solution:
Given $g\left(x\right)={x}^{\frac{1}{3}}$
Finding the derivative of the given function,
$g\text{'}\left(x\right)=\frac{1}{3}{x}^{\frac{1}{3}-1}\Rightarrow \frac{1}{3}{x}^{-\frac{2}{3}}(\because \frac{d}{dx}{x}^{n}=n{x}^{n-1})$
Again differentiating the function to find out the second derivative,
$g\text{'}\text{'}\left(x\right)=\frac{1}{3}(-\frac{2}{3}{x}^{-\frac{2}{3}-1})\Rightarrow -\frac{2}{9}{x}^{-\frac{5}{3}}$
Now, checking for the sign of g''(x)
For x=0 the function g''(x) is not defined but for $x>0;g\text{'}\text{'}\left(x\right)0andforx0;g\text{'}\text{'}\left(x\right)0$
Thus g(x) changes its concavity at x=0, hence it is an inflection point for $g\left(x\right)={x}^{\frac{1}{3}}.$
Question 1: Will the point at which f''(x)=0 exist be an inflection point for the function f(x) ?
Answer: No; If there exists an inflection point on the curve then at that point f''(x)=0 but the converse is not always true.
Question 2: Is the inflection Point a critical point?
Answer: No, the critical point is the point of local minima or local maxima while an inflection point is a point where the curve changes its concavity.
Question 3: How is the Inflection Point related to Real life scenarios?
Answer: Business growth in real life can show the inflection point as shown below.
Question 4: Can a point not lying on a given curve be an inflection point of that function?
Answer: No, the inflection point is always a point lying on the graph of the given function.