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.

f(x) = { 2x + tan x, x ≥ 0

x², x < 0

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says determine whether the following piece-wise function is differentiable at X equal to 0. And we're given the function G of X is equal to 4 X plus sine of X, if X is less than 0 and 2 X squared if X is greater than or equal to 0. We're given two possible choices as our answers. For choice A, we have yes, and for choice B, we have no. So we need to determine whether our function here is differentiable at X equal to 0. And we call that for a function to be differentiable at a given point, our function must first be continuous at that point. So we're going to check for continuity at X equal to 0 for a function G of X. So we call for a function to be continuous at a given point. The, the left-handed limit has to be equal to the right-hand limit as X approaches that point, and those two limits must be equal to the value of the function at the given point. So in our case here, we're going to have the left-hand limit, which is the limit. As X approaches 0 from the left of G of X, this must be equal to the right-handed limit, which is the limit. As X approaches 0 from the right of G of X. And that has to be equal to G of 0, the value of our function when X is equal to 0. So we need to evaluate all three of these quantities. We'll start off by looking at G of 0. If we look at our function G of X, the value of X equal to 0 is contained in the domain of our second piece, which is 2 X2, so we'll use it as our function 4 G of X. We'll have 2 multiplied by 024 G of 0, and this is going to be equal to 0. Next, we'll look at the left-handed limit, so we'll have the limit. As X approaches 0 from the left of G of X, and for G of X, we're going to have our value of X be slightly less than 0, so we're gonna want to use the first piece of our piece wise functions. So we have the limit as X approaches 0 from the left of the quantity of 4 X plus sine of X. And we can evaluate this limit by direct substitution, so this is going to be equal to 4 multiplied by 0, plus sign. Of 0. And evaluating this expression, this is going to be equal to 0. And finally, we'll look at the right-handed limit, so we'll have the limit. As X approaches 0 from the right, and here we'll be using the second piece of our piecewise function 4 G of X, so we'll have the quantity of 2 X squared. And again, we can evaluate this limit by direct substitution, so this is going to be equal to 2, multiplied by 0 squared. Which is equal to 0. And so here we see that our left-hand limit is indeed equal to our right-handed limit, and that those two limits are also equal to G of 0. So, our function is continuous at X equal to 0. So now we need to look to see if our function is differentiable. Since we've now established that our function is continuous at this point. Now for the functions to be differentiable, the left-handed derivative must also equal the right-handed derivative at the given point. So, here we're going to look at the left-handed derivative first. When X is equal to 0. And for this, we'll just use our limit definition for a derivative, so we'll have the limit. As H approaches 0 from the left. Of the quantity, and here we'll have in the numerator, G of 0 plus H. Minus G of 0. And that whole quantity is divided by H. So this is going to be equal to the limit, as H approaches 0 from the left. And for the first term of the numerator, we'll have GF H, and here H is approaching 0 from the left, so this could be um a, a value that is slightly less than 0. And so that means if we look at our function G of X, we're going to want to use the first piece of our piece-wise function, since X in that case is going to be less than 0. So, in the numerator, the first term is gonna be 4 H plus sign of H. Then we'll have minus G of 0, but we've already calculated that value, and that's going to be 0, so we'll have minus 0 in the numerator, and then all of that is divided by H. We can simplify this expression, and so this will be the limit as H approaches 0 from the left of 4. Plus, sign of H divided by H. And recall that the limit as X approaches 0 of sin of x divided by X is equal to 1. So, evaluating this limit, this is going to be equal to 4 + 1, which is equal to 5. So now we can look at the right-handed derivative. And here we're gonna be looking at the limit. As H approaches 0 from the right of G of 0 plus H. Minus G of 0, and that quantity is divided by H. So this is going to be equal to the limit, as H approaches 0 from the right. And for the first term of the numerator, we'll again have G of H, but here, H is going to be slightly greater than 0, so we're gonna need to use the second piece of our piece-wise function, which is 2 X2. So we'll have 2 H2d for the first term in our numerator, then we'll have minus G of 0, which again is 0, and that whole quantity is divided by H. Again, simplifying this expression, this is going to be equal to the limit. As H approaches 0 from the right. Of 2 H. And we can evaluate this by direct substitution. And so this is going to be equal to 2 multiplied by 0, which is equal to 0. And so here we see that our left-handed derivative is not equal to our right-handed derivative. Since 5 is not equal to 0. So since our left-handed derivative is not equal to our right-handed derivative, that means our derivative does not exist. So, that means the answer to this problem is no. And if we compare our answer here, To our answer choices, we see that the correct answer choice corresponds to answer choice B. So I hope that this explanation has been useful, and I'll see you in the next video.

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

f(x) = { 2x + tan x, x ≥ 0
x², x < 0

Key Concepts

Piecewise Functions

Differentiability

Continuity

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