If f(x)= { e^(-x)+2, for x<0
{ ax+b, for x=>0
is differentiable at x=0, then a+b=_____?
My work
lim f(x) =3
x->0-
lim f(x) =b
x->0+
3=b
lim f'(x)=-1
x->0-
lim f'(x) =a
x->0+
a=-1
therefore a+b=2
AM I RIGHT?
Please help
looks good to me.
To determine if the function f(x) is differentiable at x = 0, we need to check if the two pieces of the function meet smoothly at x = 0.
First, let's consider the right-hand side of the function, which is defined for x ≥ 0. Let's find the derivative of the right-hand side of the function with respect to x.
f(x) = ax + b
f'(x) = a
Since the derivative of the right-hand side of the function is a constant (a) and does not depend on x, it is differentiable at all points, including x = 0.
Next, let's consider the left-hand side of the function, which is defined for x < 0. Here, the function is given by:
f(x) = e^(-x) + 2
To find the derivative of this function, we can use the chain rule:
f'(x) = (-1) * e^(-x) * (-1) = e^(-x)
The derivative of the left-hand side of the function is e^(-x), which is also a continuous function. Therefore, it is differentiable at all points, including x = 0.
Since both sides of the function are differentiable at x = 0, we can conclude that the entire function f(x) is differentiable at x = 0.
To determine the value of a + b, we can substitute x = 0 into the right-hand side of the function:
f(0) = a(0) + b
f(0) = b
Therefore, a + b = a + f(0)
Since the question does not provide any specific values for a or b, we cannot determine the exact value of a + b without more information.