Inflection Point Inflection Point Today we are going to discuss about the very important topic Inflection Point Second Derivative, which you need to study in calculus. First we must have a very good knowledge of inflection point, the points where the concavity of a function changes is called as inflection point. There are two terms in which we have to deal one is concave up and other is concave down. As the name suggests concave up refers to positive second derivative and concave down refers to negative second derivative. It may happen that our function get changes from concave up to concave down and concave down to concave up. The second derivative of the given point must be equal to zero for inflection point. But we also need to take care of the things that at inflection point concavity of the function must change. We can also represent concavity in mathematical terms as, Function g(x) is concave up if and only if when it is in interval ‘I’ and all tangents on ‘I’ are below the function g(x).
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Function g(x) is concave down if and only if when it is in interval ‘I’ and all tangent on ‘I’ are above the function g(x). If we are having a point let say x = s then ‘s’ is called as inflection point if and only if the function is continuous at that point and concavity of the function changes with this point. Now we will see some examples related to concavity of the function which will clear all your doubts about the inflection point. Example1: Determine the concavity of f(x) = sin x + cos x on [0, π] and identify any points of inflection of f(x)? Solution: f(x) = sin x + cos x on [0, π], First we will calculate the first derivative of the function, f’(x) = cosx – sinx, Now we will find the second derivative of the function, f’”(x)= -sinx – cosx, We will put the value of second derivative equal to zero for finding the possible inflection point. f’”(x)=0, -sinx - cosx=0, sinx= -cosx, In the interval between 0 to π there is no point where sinx is equal to – cosx so inflection point is possible.
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For testing all intervals to the left and right of these values for f″(x) = −sin x - cos x, we find that f″(x) <0 on 0, π/2 f″(x) >0 on π /2, π Therefore function is concave downward on [0, π/2] and concave upward on (π/2, π) . Example2: f (x) = 5x3. Find the point of inflection. Solution: f (x) = 5x3 , Firstly we will calculate the first derivative of the function, f’(x) = 15x2, Now we will find the second derivative of the function, f″(x) = 30x, Then the second derivative is f’”(x) = 30x. Now set the above second derivative function equal to zero and solve for "x" to find possible inflection points. 30x=0, x=0, And if x=-1, 1 than, F "(-2) = -60, f "(2) = 60. The inflection point is at x=0 where the function has different concavities on either side., in many cases we can see that inflection point but in cases there is no inflection point but function still goes under concave up and concave down. In this way we can solve the question related to concavity.
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