- Thread starter
- #1
Welcome to our community
Be a part of something great, join today!
differentiation
- Thread starter Yankel
- Start date
- Admin
- #2
- Thread starter
- #3
- Admin
- #4
- Feb 15, 2012
- 1,967
checking for continuity is a good idea, but in this particular problem it doesn't rule out any points for us, f is continuous everywhere:
$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 2 = 2$$
$$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{2}{x^2} = 2$$
so f is continuous at 1.
$$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{2}{x^2} = \frac{2}{4} = \frac{1}{2}$$
$$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \frac{1}{x} = \frac{1}{2}$$
so f is continuous at 2.
there's no problem at 0, f(0) = 2.
instead, you should look at places where the derivative might not exist.
$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 2 = 2$$
$$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{2}{x^2} = 2$$
so f is continuous at 1.
$$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{2}{x^2} = \frac{2}{4} = \frac{1}{2}$$
$$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \frac{1}{x} = \frac{1}{2}$$
so f is continuous at 2.
there's no problem at 0, f(0) = 2.
instead, you should look at places where the derivative might not exist.
Last edited:
- Jan 29, 2012
- 1,151
Since the question asks about "differentiability" I would think that the obvious first thing to do would be to look up the definition of "differentiable". Fortunately, for a function of a single variable that is simply "has a derivative there" (for functions of more than one variable it is more complicated).
So what is the definition of "the derivative of f(x) at x= a"?
(You are completely wrong when you say the function is not continuous. I think you need to review the basic definitions.)
- Admin
- #7