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.
To determine if the function g(x) is differentiable at x = 0, we need to make sure that the left-hand derivative and the right-hand derivative match at that point.
First, let's find the left-hand derivative of g(x) at x = 0. For x < 0, g(x) = x + b.
Taking the derivative of g(x) = x + b with respect to x, we get:
g'(x) = 1, for x < 0.
Next, let's find the right-hand derivative of g(x) at x = 0. For x ≥ 0, g(x) = cos(x).
Taking the derivative of g(x) = cos(x) with respect to x, we get:
g'(x) = -sin(x), for x ≥ 0.
Now, let's evaluate the derivatives at x = 0:
For the left-hand derivative, since x < 0, we substitute x = 0 into g'(x) = 1:
g'(0-) = 1.
For the right-hand derivative, since x ≥ 0, we substitute x = 0 into g'(x) = -sin(x):
g'(0+) = -sin(0) = 0.
Since g'(0-) = 1 and g'(0+) = 0, the left-hand derivative and right-hand derivative do not match at x = 0. Therefore, the function g(x) is not differentiable at x = 0 for any value of b.
Hence, your answer of 0 will not make the function g(x) differentiable at x = 0.