Question

F(x) = {x^2 ,⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡−⁡⁡⁡1 ≤ x < 0 {-x, 0 ≤ x ≤ 1 a. Is ƒ continuous at x = 0? b. Is ƒ differentiable at x = 0? Give reasons for your answers.

Answer 1

Think you mean this:

• `f(x) = {(x^2 ,qquad ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡−⁡⁡⁡1 ≤ x < 0),( -x,qquad 0 ≤ x ≤ 1):}`

Looks like this:

graph{(y - x^2)(y+x) = 0 [-0.195, 0.2153, -0.089, 0.1162]}

a) Continuity

This simplest test for continuity is: Can you swoop in from the left along `y = x^2` to the Origin, then change direction and follow `y = - x` from there, without lifting your pen from the paper?

If so, the function is continuous at the Origin. And, from the plot, clearly you can.

You can refine that by stating that `f(x)` is continuous at `f(0)` provided that:

• `{(f(0) " is defined"),( lim_(x to 0) f(0) = f(0)):}`

That second test requires the right-sided limit, applied to `y = x^2`, to agree with the left-sided limit, applied to `y = - x`. Here, they agree.

b) Differentiability

At `x = 0` there is an abrupt change in direction. Yet, for a function to be differentiable , the limit has to be 2-sided.

Here, it is not diferentiable.