Find the value of b, if any, that will make the function differentiable at x = 0

g(x)= { x+b, x<0
cos(x), x≥0

can you help, I got 0 but I don't know if that will make the function differential at x=0.

On APEX the answer is: No such value exists

I'm pretty sure this is because for any value you put for b the graph of the function at x=0 is either discontinuous (so it can't be differentiable) or forms a corner, which basically means the function does not approach the same tangent line at the corner from the left- and right-hand sides. Therefore the function is not differentiable at that point.

Hope this was helpful

Well, finding the value of b that makes the function differentiable at x=0 requires a little bit of calculus and a little bit of clowning around! Let's investigate.

For a function to be differentiable at a certain point, its derivative must exist at that point. In this case, we need to find the derivative of g(x) and evaluate it at x=0.

For x < 0, the derivative of g(x) is simply 1, because the derivative of x+b with respect to x is just 1.

For x ≥ 0, the derivative of g(x) is -sin(x), because the derivative of cos(x) with respect to x is -sin(x).

Now, let's evaluate the derivatives at x=0. For x < 0, the derivative is 1. For x ≥ 0, the derivative is -sin(0) = 0.

Since the derivative exists and is equal to 1 on one side of x=0, and exists and is equal to 0 on the other side, the function is not differentiable at x=0.

So, the value of b doesn't matter in this case. No matter what value you choose for b, the function will not be differentiable at x=0.

But hey, don't be sad! There's always another equation waiting to be solved and another laugh waiting to be had.

To determine the value of b that will make the function g(x) differentiable at x = 0, we need to check if the left-hand limit and right-hand limit of g(x) as x approaches 0 are equal.

First, let's find the left-hand limit:

lim (x → 0-) (x + b)

Since x approaches 0 from the left, we can simplify the expression to:

lim (x → 0-) x + b = 0 + b = b

Now, let's find the right-hand limit:

lim (x → 0+) cos(x)

Since cos(x) is a continuous function, we can evaluate the limit directly:

lim (x → 0+) cos(x) = cos(0) = 1

For g(x) to be differentiable at x = 0, the left-hand limit should be equal to the right-hand limit.

b = 1

Therefore, the value of b that will make the function g(x) differentiable at x = 0 is b = 1.

To determine if the function g(x) is differentiable at x = 0, we need to check if both pieces of the function are differentiable at that point and if the values of the function and its derivative are continuous at x = 0.

First, let's check if the function x + b is differentiable at x = 0 when x < 0.

For differentiability, we need to check if the limit of the difference quotient exists as x approaches 0 from the left:

lim(x→0-) [(x + b) - (0 + b)] / (x - 0)

Simplifying, we get:

lim(x→0-) (x + b) / x

Now, calculate the limit by substituting x = 0:

lim(x→0-) (0 + b) / 0

Since the denominator approaches 0 from the left side, the limit does not exist. Therefore, the function x + b is not differentiable at x = 0 when x < 0.

Next, let's check if the function cos(x) is differentiable at x = 0 when x ≥ 0.

The function cos(x) is a trigonometric function, and it is known to be differentiable for all real values of x. Therefore, the function cos(x) is differentiable at x = 0.

Now, to determine if the function g(x) is differentiable at x = 0, both pieces of the function must be differentiable at that point. Since x + b is not differentiable at x = 0 when x < 0, there is no value of b that will make the function g(x) differentiable at x = 0.

Hence, the function g(x) is not differentiable at x = 0 regardless of the value of b.