For what value or values of the constant m, if any, i...

Question

For what value or values of the constant m, if any, is

ƒ(x) = { sin 2x, x ≤ 0

{ mx, x > 0

a. continuous at x = 0?

b. differentiable at x = 0?

Give reasons for your answers.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says for what value or values of the constant K, if any, is the function G of X defined as G of X, which is equal to sin of X if X is less than 0, and K multiplied by X2 if X is greater than or equal to 0, continuous at x equal to 0, and differentiable at X equal to 0. So for this problem, we need to find the values of K that make our function continuous at X equal to 0 and differentiable at X equal to 0. So we'll start off by looking at the continuity first. So we call that for a function to be continuous at a point, that the left-hand limit and the right hand limit need to be equal to each other, and they also need to be equal to the value of the function at that point. So in our case, we're gonna need the limit. As X approaches 0 from the left of G of X. To be equal to the limit as x approaches 0 from the right of G of X. Which is equal to G of 0. So we'll start off by calculating the value of G of 0. So, for G of 0, we look at our function G of X, when X is equal to 0, we want to use the piece that is K multiplied by X2. So we'll have K multiplied by 0 squared, which is equal to 0. So now we can look at the left-handed limit. So we'll have the limit, as X approaches 0 from the left, and for G of X, since X is going to be slightly less than 0, we're going to use the portion that is sin of X. So we'll have the limit as X approaches 0 from the left of sin of X. And we can evaluate this limit by direct substitution, so this will be a sign of 0, which is equal to 0. And lastly, we'll look at the right-handed limits. We'll have the limit as X approaches 0 from the right of G of X and for G of X, we're going to use the portion that is K multiplied by X2, since X is going to be slightly larger than 0 as it approaches 0 from the right. And so we'll have the limit as X approaches 0 from the right of K X squared. Again, we can use direct substitution to evaluate this limit, so this is going to be equal to K multiplied by 0 squad, which is equal to 0. So, if we look, all three of our expressions here are equal to 0. And notice that this is for any value of K. So that means that our function is continuous. Or Guinea. Value of K. So that is going to be the answer to the first part of our problem, that our function is continuous for any value of K. So now we can look at differentiability. And for a function to be differentiable, we need both the left hand derivative and the right hand derivative to be equal to each other. So, we're gonna look at the limit. As X approaches 0 from the left. Of G of X. And this is going to be equal to the limit. As X approaches 0 from the left. And here for G of X, since X is going to be slightly less than 0, we're going to want to use sine of X or G of X. So G of X is going to be the derivative with respect to X. Of sine of X. And this is going to be equal to the limit, as X approaches 0 from the left. And when I take the derivative of sin of X with respect to X, I get cosine of X. And I can evaluate this limit by direct substitution, but this is going to be equal to the cosine of 0, which is equal to 1. Next, we'll look at the right-hand limit, or the right-hand derivative. So, we'll have the limit as X approaches 0 from the right of G of X. And this is going to be equal to the limit, as X approaches 0 from the right. Of the derivative with respect to x. Of the quantity of K multiplied by X squared in quantity. Since X is gonna be slightly larger than 0, we'll want to use the um piece of K multiplied by X2d as our function for G of X. So we can use our power rule to calculate this derivative. This is going to be equal to the limit, as X approaches 0 from the right of 2 multiplied by K multiplied by X. And we can evaluate this limit by direct substitution, so this is going to be equal to 2 multiplied by K multiplied by 0. And evaluating that expression, this is going to be equal to 0. So notice that For our left limit and right limit that I have one for the left limit, but that is not equal to 0, which is the value of the right limit. So, here, no matter what value of K I have, I'm always gonna have 1, not equal to 0. So, when we look at differentiability, Our function is Differentable For no value. Of K. That's because one is never going to be equal. To 0, regardless of the value of K. So that is going to be the second half of our question answered. So just to recap, we found that our function was continuous for any value of K, but it was differentiable for no value of K. So that is going to be the answer to this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

For what value or values of the constant m, if any, is

ƒ(x) = { sin 2x, x ≤ 0
{ mx, x > 0

a. continuous at x = 0?
b. differentiable at x = 0?

Give reasons for your answers.

Key Concepts

Continuity

Differentiability

Piecewise Functions

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