Is there a value of b that will make

g(x) = {...

Question

Is there a value of b that will make

g(x) = { x + b, x < 0

cos x, x ≥ 0

continuous at x = 0? Differentiable at x = 0? Give reasons for your answers.

Accepted Answer

Welcome back everyone. Find the value of C that ensures the function G of X is a continuous at X equals 0 and B differentiable at X equals 0. We're given our piece wise function G of X, which is defined as X plus C if X is less than 0 and 2 cosine of 2 x if X is greater than or equal to 0. So let's begin with the first part that would be Part A. We want to check continuity. According to the continuity condition, we want to ensure that limit as X approaches 0 from the left, right? We're looking at X equals 0 of G of X is equal to the limit. As x approaches 0 from the right of G of X and those must be also equal to the value of the function at the point of interest, that would be 0, right? So let's evaluate each limit. Limit as x approaches 0 from the left. We're going to take the first part of the piece wise function because we're looking at X is less than 0, so we're taking the limit of X plus C. We can perform direct substitution. So X is 0, we simply get C, right? Now for our next limit, we get limit as X approaches 0 from the right. We're going to take to cosign. Of 2 X, and we can substitute X equals 0. Cosine of 0 is 1, so we simply get a limit of 2. And LG of 0 is basically equal to that limit, right, because our function is defined for X equals 0 at 2 cosine of 2 X. So what we want to do is simply equate 2 equals C, which tells us that C equals 2 is the value that we're looking for to make our function continuous at X equals 0, right? And now let's move on to the second part of the problem, that would be part B. What we want to do is simply evaluate the derivative of G of X. And we're going to find the derivative of each part, so we begin by differentiating X plus C. If X is less than 0, and then we're going to differentiate to cosine of 2 X if X is greater than or equal to 0. Let's perform differentiation, this gives us. Now, the derivative of X is one. The derivative of C is 0 because it's a constant, so the derivative is 1 if x is less than 0, and now the derivative of cosine is negative sine. So we get -2 sin of 2 X for the second part of the derivative. And now, according to the chain rule, we also have to multiply by the derivative of the inner function and the derivative of 2 X is 2. So we get -4 of 2X. If x is greater than or equal to 0. Now that we have each derivative, what we want to do is simply evaluate each limit, right? So we're going to set a limit. As x approaches 0 from the left. Of G of X must be equal to the limit as X approaches 0 from the right of G of X. This is the differentiability condition. We want to ensure that each slope is exactly the same if we approach that point from the left and from the right. So we want to show that limit. As X approaches 0 from the left. Of G of X, which gives us 1, right? X is less than 0. This gives us a value of one. is equal to the limit as x approaches 0 from the right of -4 of 2 X. We can perform direct substitution. And show that we want to ensure that 1 is equal to -4 of 2 X. Now, is this a valid equation? Well, we can actually solve it and identify. A set of x values which satisfies the equation, but apparently not every X will satisfy it and also this equation is independent of C, right? And because it's independent of C, we can essentially say that there is no value of C that can make G of X differentiable at X equals 0, right? Specifically. If we substitute X equals 0, this means that -4 sin of 0, which is 0, gives us an equation 1 equals 0, right? And this is invalid. Therefore, Because 1 is not equal to 0, we can say that there is no c value that can make our function differentiable at X equals 0. Therefore, for a, we can say that C equals 2, and for part B, we can say that there is no C. That makes. It Differentiable That's it. We have our answers and thank you for watching.

Is there a value of b that will make

g(x) = { x + b, x < 0
cos x, x ≥ 0

continuous at x = 0? Differentiable at x = 0? Give reasons for your answers.

Key Concepts

Continuity

Differentiability

Piecewise Functions

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