Continuity of a Function

First we have to prove that if a function $f$ satisfies the condition $f(x+y)=f(x)+f(y)$ for all real values of $x$ and $y,$ then $f(0)=0$ . Indeed, making $x=y=0$ in the condition, we obtain that $f(0)=f(0)+f(0)$ and this implies that $f(0)=0.$

Let $a$ be an arbitrary real number. We can prove that $f$ is continuous at $a.$ Making $x=a$ and $y=t-a$ in the given functional equation, we get that $f(t)=f(a)+f(t-a)$ for all real value of $t$ . Finding limits of both sides as $t$ tends to $a$ we get $\lim_{t\rightarrow a} f(t)= \lim_{t\rightarrow a} f(a)+\lim_{t\rightarrow a} f(t-a)=f(a)+\lim_{z\rightarrow 0} f(z)$ $=f(a)+f(0)=f(a)+0=f(a)$

So $f$ is continuous at $a.$ This proves that $f$ is continuous at any number $a.$

Hit and trial:- simply f(x) is a linear function every linear function is statistics it. And a linear function is always continuious and differenciable.

f(x) = kx is the function which satisfies the relation.

For proof differentiate the function partially with respect to x.

Then, f'(x) = f'(0).

f'(0) is a constant. Let us take it as k.

Now, integrate on both sides.(/ is integral).

/f'(x) = /k implies,

f(x) = kx + c where c is integration constant.

if x = 0, c = f(0).

Now to find f(0), put x=0 and y=0 in the given relation.

Then, f(0) = 2f(0) implies f(0) = c = 0.

Thus f(x) = kx satisfies the given relation.

It is a straight discontinuous line passing through origin with slope = k.

So, number of discontinuity points in f(x) is 0.