Determine whether the function is differentiable at x=2
x^2+1 for x<(or equal to) 2
4x-3 for x>2
I did the math for the limits of both equations and they both approach to 4. So that means they are differentiable right?
Any ideas?
for the function to be continuous, the limit on both sides must be the same. They are both 5 (not 4!).
So, your function is continuous. But that is not enough. Think of f(x) = |x|. It is continuous, but not differentiable at x=0.
For it to be differentiable, the derivative on both sides must exist and have the same limit. For your function, the derivatives are
left: 2x
right: 4
2x=4 at x=2, so it is differentiable. That means it is in some sense "smooth" where the pieces meet:
http://www.wolframalpha.com/input/?i=plot+y%3Dx%5E2%2B1,+y%3D4x-3,+0+%3C%3D+x+%3C%3D+4
You can see that they fit smoothly together at x=2.
To determine whether a function is differentiable at a particular point, we need to check if the left-hand derivative and the right-hand derivative exist and are equal at that point.
In this case, let's analyze the function in two parts:
For x β€ 2, the function is given by f(x) = x^2 + 1.
For x > 2, the function is given by f(x) = 4x - 3.
First, let's find the derivative of each part:
For x β€ 2: f'(x) = 2x. (Derivative of x^2 + 1)
For x > 2: f'(x) = 4. (Derivative of 4x - 3)
Now, let's determine the left-hand derivative (f'(2-)) and the right-hand derivative (f'(2+)):
For the left-hand derivative:
f'(2-) = limit as x approaches 2 from the left of f'(x) = limit as x approaches 2 from the left of 2x = 2 * 2 = 4.
For the right-hand derivative:
f'(2+) = limit as x approaches 2 from the right of f'(x) = limit as x approaches 2 from the right of 4 = 4.
Since f'(2-) = f'(2+), which is equal to 4 in both cases, the function is differentiable at x = 2.
Therefore, your conclusion that the limits of the derivatives approaching 4 means the function is differentiable at x = 2 is indeed correct.