Points of inflexion. Hello friends, today it’s all about the points of inflexion. Have a look!!

**Points of inflexion**

**Solved examples of the points of inflexion**

Disclaimer: None of these examples are mine. I have chosen these from some book or books. I have also given the due reference at the end of the post.

So here is the first example.

**Example 1**

According to Stroud and Booth (2013), “Determine the smallest positive value of at which a point of inflexion occurs on the graph of .”

**Solution**

Now here the equation of the curve is

(1)

And I have to get the point of inflexion. So I’ll start with the differentiation.

**Step 1**

First of all, I’ll differentiate equation (1) with respect to . Since the function is a product of two functions and , I’ll use the * product rule of differentiation* to differentiate . Thus it will be

Now I’ll simplify it to get

(2)

Next, I’ll differentiate equation (2) with respect to . Again I’ll use the same product rule of differentiation as before. So I get

Then I’ll simplify it. So this gives

Hence I can say that

(3)

Now, for any point of inflexion, the value of is equal to . Since for any point of inflexion , I’ll solve equation (3). And that gives

So either or .

Now for . And for . So that means which gives .

Next, I’ll check if there is a point of inflexion at . If there is a point of inflexion at , there will be a change of sign at as we pass through it.

**Step 2**

So, let’s choose a point which is a little bit higher than say, . Now I’ll substitute this value in equation (3) to get

And that means

Since is very small, is less than . Thus is less than , that is, negative. Also, we all know that . Thus will have a negative sign.

So the sign of will be

Now I’ll choose a point which is a little bit lower than say, . Next, I’ll check the sign of for .

Then I’ll substitute in equation (3) to get

And that means

So that gives

Since has a very small positive value, is always positive. So is also positive. And is always positive. Thus the sign of for is which is positive.

As I can see, there is no change of sign in as it passes through . Thus there is no point of inflexion at .

Next, I’ll check it for .

**Step 3 **

So, let’s choose a point which is a little bit higher than say, where is a very small number.

Now I’ll substitute this value in equation (3) to get

And that means

Since is positive, is also positive. Thus will have a positive sign. So the sign of will be

Now I’ll choose a point which is a little bit lower than say, . Next, I’ll check the sign of for .

Now I’ll substitute in equation (3) to get

And that means

Since is positive, is also positive. Also, we all know that . Thus will have a negative sign.

So the sign of will be

As I can see, there is a change of sign at as it passes through . So there is a point of inflexion at . Since there is no other point of inflexion, I can say that is the lowest value of at which the point of inflexion occurs.

Hence I can conclude that this is the answer to the given example.

Next comes my other example.

**Example 2**

According to Stroud and Booth (2013), “Find the coordinates of the points of inflexion on the curves .”

**Solution**

Now here the equation of the curve is

(4)

And I have to get the coordinates of the points of inflexion. So I’ll start with the differentiation.

**Step 1**

First of all, I’ll differentiate equation (4) with respect to . Thus it will be

Now I’ll simplify it to get

Again I’ll differentiate it with respect to to get

(5)

Now, for any point of inflexion, the value of is equal to . So here comes my next step.

**Step 2**

Since for any point of inflexion , I’ll solve equation (5). And that gives

So that means

So for any point of inflexion, there is a change of sign at as it passes through that point.

Now I’ll choose a point which is a little bit lower than say, . Next, I’ll check the sign of for .

Then I’ll substitute in equation (5) to get

And that means

Next, I’ll choose a point which is a little bit higher than say, . Then I’ll check the sign of for .

So I’ll substitute in equation (5) to get

And that means

As I can see, there is a change of sign at as it passes through . Ao is the x-coordinate of the point of inflexion. Now I’ll get the y-coordinate.

Hence I’ll substitute in equation (4) to get the value of as

Next, I’ll simplify it. So that means

Thus are the coordinates of the point of inflexion of this curve.

Hence I can conclude that this is the answer to the given example.

Dear friends, this is the end of today’s post. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!

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