# differentiation

#### MarkFL

Staff member
I would first check for continuity. Is the function continuous over the entire domain?

#### Yankel

##### Active member
well it ain't, there is a problem with x=1,2 and I think also x=0

#### MarkFL

Staff member
Can you state why you think there are problems at the x-values you give?

#### Deveno

##### Well-known member
MHB Math Scholar
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.

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#### HallsofIvy

##### Well-known member
MHB Math Helper
Hello,

How do I solve this kind of problems ?

For which values of x the next function is "differentiable" ?

View attachment 530

I know it has something to do with the existent of the one sided limits, but which limits should I be calculating exactly ?

Thanks !
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.)

#### MarkFL

$\displaystyle \lim_{x\to1^-}f'(x)=\lim_{x\to1^+}f'(x)$
$\displaystyle \lim_{x\to2^-}f'(x)=\lim_{x\to2^+}f'(x)$