# Antiderivative of a function and intermediate value property

I have learnt that if a function does not satisfy the intermediate value property (IVP), then its antiderivative does not exist. For instance, $f(x)= [x]$, where $[\cdot ]$ denotes greatest integer function, has no antiderivative as it doesn't satisfy IVP. In the same manner, how can we say $f(x)= \frac{\sin x}{x}$ has no antiderivative? Also, I am confused about whether a function $f(x)$ satisfies IVP or not. And what about $g(x)= \exp(- x^2)$?

The function $f(x)=\frac{\sin x}{x}$ is a continuous function, so it does have an antiderivative.

Same with the function $g(x)=e^{-x^2}$.

The two antiderivatives, however, are not elementary, which means they cannot be written in closed form as a combination of trigonometric functions, polynomials and exponentials (and their sums, product, divisions etc).

So, we cannot say that the functions you mention "do not have an antiderivative" because that's just not true. We can only say that the functions do not have an elementary derivative.

• And whether f and g satisfying IVP? – a learner Aug 28 '17 at 7:44
• @alearner They are continuous, so they do satisfy the IVP. – 5xum Aug 28 '17 at 7:45
• Ohh yes, sorry sir I just didn't focused on that thing. Thank you very much and please tell me is there any function which is continuous in [a,b] and differentiable in (a,b) but not differentiable at boundary point a and b ? – a learner Aug 28 '17 at 7:51
• @alearner $\sqrt{1-x^2}$ on $[-1,1]$ is a good example. – 5xum Aug 28 '17 at 7:51
• Yes sir then darbaux principle says, Let I be a closed interval, f : I ----> R a real-valued differentiable function. Then f ' has the intermediate value property: If a and b are points in I with a < b , then for every y between f '(a ) and f '(b) , there exists an x in (a,b) such that f'( x )= y. So I have doubt that if we have function which is differentiable on open interval I ( not closed) then we can't use darbaux principle. Right? – a learner Aug 28 '17 at 8:03