Question

Dear kind-hearted tutor,
Can you check my solution on this question?
Given f' (x) = {2x - 1, x≤1; x^2, x>1, find f’ (1) if exist.

Since it asks to find f’ (1). I choose the first derivative because it satisfies the condition for 1.
2 x – 1 = 2 (1) – 1
= 1

What I am unclear is how to show that it exist? Would you please give me some clear idea on this question? Your help is very much appreciated.

Thank you.

Answer 1

It says f'(x) = 2x-1 for x ≤ 1,
so x = 1 is included and valid for f'(x) to get f'(1) = 1

for x ≤ 1, f(x) = x^2 - x + c
which is a parabola opening upwards, we don't know c
but notice that when x = 1 , f(1) = c
so it "ends" at (1,c)

for x> 1, f(x) = (1/3)x^3 +k , which is a cubic and again we don't know the value of the constant k,
but we have a cubic which "begins" at (1, 1+k)

I graphed both parts of the function with constants of 0 in both cases.
Look at the parabola (blue) for only x ≤ 1
and look at the cubic (red) for only x > 1
http://www.wolframalpha.com/input/?i=y+%3D+x%5E2+-+x+,+y+%3D+(1%2F3)x%5E3

Adding constants would only shift each curve vertically.
The key point is that x = 1 is part of the parabola.