In Exercises 41–44, determine whether the piecewise-defined fu...

Question

In Exercises 41–44, determine whether the piecewise-defined function is differentiable at x = 0.

g(x) = { 2x − x³ − 1, x ≥ 0

x − (1 / (x + 1)), x < 0

Accepted Answer

Hello. In this video, we are going to be checking if the given function below is differentiable at the value X is equal to 0. We are given a piecewise function. F of X is defined as sin of X when X is greater than or equal to 0, and the function is defined as X2, multiplied by cosine of 1 divided by X, when X is less than 0. In order to check if a function is differentiable, the first thing we want to check is the continuity of the function. Is the function continuous? So, in order to check the continuity of a function, we first need to check the left hand and right hand limit of the piece wise function. Let's go ahead and start with the left hand limit. We are going to take the limit as X approaches 0 from the left. Now, when we are approaching 0 from the left, we are dealing with values of X that are less than 0, and when X is less than 0, the piecewise function is defined as X2, multiplied by cosine of 1 divided by X. Now, in this case here, if we directly substitute the value of 0, we will get an undefined value, since we have division by zero in the cosine function. So we need to figure out another approach in order to determine the limit of this function. Well, one approach we can take is the squeeze theorem. We can simplify this function further, but we do know that the maximum and minimum value of cosine is going to be -1 and 1. So we can bound our cosine function between -1 and 1. Now what we want to do is we want to go ahead and multiply this inequality by the terms that we are missing in the original function. We are missing the X2 term, so we are going to multiply the entire inequality by X2. That is going to leave us with negative X2 less than or equal to X2 multiplied by cosine of 1 over X, less than or equal to X2. And now what we want to do is we want to go ahead and take the limit. As X approaches 0 from the left of the endpoints of this function. Now the limit as X approaches 0 of negative X squared is going to give us 0. X2 multiplied by cosine of 1 over X will be greater than this value. And when we take the limit, as X approaches 0 of X2, we get the value of 0. Notice that the function X2 multiplied by 1 divided by X is bounded between the values of 0. What this means is that the left hand limit is going to have the value of 0 by the squeeze theorem. Now we need to check the right hand limit. We will check the limit as X approaches 0 from the right. We are working with values of X that are greater than 0, and in this case here, the function is going to be defined as sign of X. If we directly substitute the value of our limit, we will have sign of zero, which is equal to 1. Oh, sorry, sin of 0 equal to 0. So, what this means is that the function is continuous. But now that we've figured out that the function is continuous, the next thing we want to check is to see if the left hand and right hand derivatives. Are equal to each other. So, in order to check the left hand and right hand derivatives, we are going to use the definition of the derivative. Now, recall that the definition of the derivative is given to us as the limit as H approaches 0 of F A plus H minus F of A is divided by H. Now, in this case here, because we want to check the derivatives around the value of 0, we are going to let A equal to 0. So we have the limit, as H approaches 0, of F 0 plus H, which is F of H, minus F of 0 is divided by H. So, for the left hand derivative, we will go ahead and set up the limit as the limit as H approaches 0. From the left. And that is going to be. H plugged into X2 cosine 1 divided by X, which will leave us with H2 multiplied by cosine of 1 divided by H minus the value of 0. Now, again, we don't know what the value of zero is, but we know that the limit of this function approaching 0 is 0. So we will replace that value with 0, and this will be divided by H. This will further simplify the derivative to be the limit as H approaches 0 of H2d multiplied by cosine of 1 divided by H, all divided by H. We can factor out an H from the numerator and denominator, allowing us to simplify the function to be the limit as H approaches 0 from the left of H multiplied by cosine 1 divided by H. And if we repeat the process of the squeeze theorem on this problem, we will get that the limit is equal to 0. So the derivative of the left-hand side of the function will be 0. Now we will check the right hand limit. We have the limit as H approaches 0 from the right of sine or H plugged into the sine function, which will be sine of H minus the value of 0 plugged into sine, which is 0, all divided by H. This will simplify to be the limit as H approaches 0 from the right of sine of H divided by H. This is in the form of the special sine limit, and this is going to simplify to be 1. Notice that the values of the left hand and right hand limits do not equal to each other. Therefore, this function is not differentiable, and this is going to be the solution to the problem. So I hope this video helps you in understanding on how to approach this type of problem, and I will go ahead and see you all in the next video.

In Exercises 41–44, determine whether the piecewise-defined function is differentiable at x = 0.

g(x) = { 2x − x³ − 1, x ≥ 0
x − (1 / (x + 1)), x < 0

Key Concepts

Piecewise-Defined Functions

Continuity at a Point

Differentiability and Derivatives

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