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 whether the function g(x) is differentiable at x = 0, we need to check if the two parts of the function (x + b for x < 0 and cos(x) for x ≥ 0) have the same value and slope at that point.
Let's start by evaluating the function at x = 0 from both sides.
For x < 0:
g(x) = x + b = 0 + b = b
For x ≥ 0:
g(x) = cos(x) = cos(0) = 1
So, for g(x) to be differentiable at x = 0, the value of g(x) should be the same from both sides, which means b must be equal to 1.
Next, let's check if the slopes (derivatives) are the same at x = 0.
For x < 0:
g'(x) = 1 (since the derivative of x + b is simply 1)
For x > 0:
g'(x) = -sin(x) (since the derivative of cos(x) is -sin(x))
At x = 0, the derivative should be continuous and have the same value from both sides. Therefore, we need to check if the limit of g'(x) as x approaches 0 from both sides is equal.
For x approaching 0 from the left (x < 0):
lim (x→0-) g'(x) = lim (x→0-) 1 = 1
For x approaching 0 from the right (x > 0):
lim (x→0+) g'(x) = lim (x→0+) (-sin(x)) = 0
Since the two limits are not equal, the function g(x) is not differentiable at x = 0, regardless of the value of b.
In conclusion, it is not possible to find a value of b that will make the function g(x) differentiable at x = 0.