Is f(x) Differentiable at x = 1?

[Miike012]

let

f(x) = 2-x if x<= 1
x^2 - 2x + 2 if x > 1

Is f diff at x = 1?

At first I would say yes because f(x) is continuous at x = 1.

But when I graph f '(x) it is obvious that the function is not differentiable at x = 1.

My questions is... is there another way to determine if f is diff at x = 1 other than graphing?

thank you.

 

[Char. Limit]

In order for a function to be differentiable, the derivative must exist at every point in the domain. For all x other than x=1, this is easy to show. However, the derivative cannot exist at x=1. To show this easily, try finding the slope as x approaches 1 from the left, then from the right. For the derivative to exist, those two slopes have to be equal.

 

[Miike012]

Thank you

 

[Willowz]

let

f(x) = 2-x if x<= 1
x^2 - 2x + 2 if x > 1

Is f diff at x = 1?

At first I would say yes because f(x) is continuous at x = 1.

But when I graph f '(x) it is obvious that the function is not differentiable at x = 1.

My questions is... is there another way to determine if f is diff at x = 1 other than graphing?

thank you.

Are you asking if the function is continuous at those points? Because if you want to find the derivative it has to first be continuous on the point. But, in this example it won't be differentiable because there is a sharp point at x=1, I think.