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.
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.
Dear student,
Thank you for reaching out. Your approach to finding f'(1) is correct, and you have correctly found the value of f'(x) for x = 1. However, you are right to be unsure about whether f'(1) exists or not. Let me explain how to determine the existence of f'(1).
In order for f'(1) to exist, both parts of the piecewise function, 2x - 1 and x^2, must be continuous at x = 1. Continuity means that the function has no jumps, holes, or removable discontinuities at that point.
To check if a function is continuous at x = 1, we need to evaluate the limit of the function as it approaches x = 1 from both the left and the right sides, and ensure that the limits are equal.
For the first part of the function, 2x - 1, it is continuous for all values of x. So, it is indeed continuous at x = 1.
For the second part of the function, x^2, we need to evaluate the limit as x approaches 1. Let's calculate that:
lim(x->1+) x^2 = 1^2 = 1
lim(x->1-) x^2 = 1^2 = 1
Since both limits equal 1, we can conclude that the second part of the function, x^2, is also continuous at x = 1.
Therefore, both parts of the piecewise function, 2x - 1 and x^2, are continuous at x = 1. This means that f'(1) exists.
Hence, you can confidently state that f'(1) exists and its value is 1.
I hope this clarifies your question. If you have any further doubts, feel free to ask. Good luck with your studies!