How can I prove that this function is differentiable at x = 1?

Question

It's a piecewise polynomial function:

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

It's a parabola that turns into a line. It doesn't have any gaps or corners. The limit of f(x) as x approaches 1 is 2, and the limit of f'(x) as x approaches 1 is 2.

This was a problem on a test, but I my calculus teacher took points off because she says that the function is not differentiable at x = 1. I need to formally prove that this function is differentiable at x = 1, so I can get those points back.

Thanks in advance!

Answer 1

The first condition for a function to be differentiable in a point x is that the function is continuous at that point.
it's easy to prove the function is continuous at x=1
because:
Lim f(x)=Lim(x^2 + 1)= 2 for x going to 1 from left (1-)
Lim f(x)=Lim(2x)=2 for x going to 1 from right (1+)
We have now proved that the function is continuous at x=1

We will look at f' at x=1
f'(x)=2x ; x<=1
f'(x)=2 ; x>1
the function is differentiable in a point x=a if and only if, the function is continuous and f' exist and has the same value when x goes to point x=a from left and right

Lim f'(x)=2 for x going to x=1 from left (1-)
Lim f'(x)=2 for x going to x=1 from right (1+)
now we have proven that the function is differentiable at x=1

Answer 2

properly, for any actual function f(f(x)) = e^-x + a million, f would be a million-a million. assume f isn't a million-a million, then there are 2 distinctive reals x,y such that f(x)=f(y), which shows that f(f(x)) = f(f(y)). => e^-x + a million = e^-y + a million => e^-x = e^-y => -x=-y so as that x=y, a contradiction. d/dx f(f(x)) = -e^-x < 0 for all x additionally, d/dx f(f(x)) = f'(f(x)) * f'(x). Now, if f is a a million-a million function and differentiable, then it is non-provide up and subsequently the two weakly increasing or weakly reducing. Case a million: assume it is weakly increasing (f'(x) >= 0 for all x): d/dx f(f(x)) = f'(f(x)) * f'(x) is the manufactured from 2 nonnegative numbers and subsequently nonnegative. yet it is a contradiction, by way of fact the derivative is destructive everywhere. Case 2: assume it is weakly reducing (f'(x) <= 0 for all x): resembling above, the place the manufactured from 2 such numbers shows that the derivative ought to be nonnegative. =><=. subsequently no such differentiable function exists. ------- surely, i think of we are able to weaken the argument interior the 2d section slightly, showing that no such non-provide up function exists. although, i don't be attentive to on the topic of the discontinuous ones. (and my gut feeling is that it would ought to be discontinuous everywhere, yet I surely have no longer something to decrease back that up)