Given that $f(a)=\int_{2}^{a} e^{u^2}du$
a) Is $f(a)$ differentiable?
b) Is $f(a)$ one to one?
c) Does $f^{-1} (0)$ exist? If yes, find the value
d) Does $(f^{-1}) '(0)$ exist? If yes, find the value
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a) $f(a)$ is differentiable if $\lim_{h\to 0} \frac{f(a+h)-f(a)}{h}$ exists but how to calculate the limit $\lim_{h\to 0} \frac{\int_{2}^{a+h} e^{u^2}du - \int_{2}^{a} e^{u^2}du}{h}$ ?
I suspect there is an easier method to check the differentiability but I can not think of one
b) Yes $f(a)$ is one to one function because different values of $a$ will give different values of the integration since the integration represents the area under graph of $e^{u^2}$ from 2 to $a$. The bigger the value of $a$, the bigger the area and if $a$ is less than 2, the result of the integration will be negative.
Is this a good enough reasoning, or is there a more rigorous way to prove it?
c) Let $f^{-1} (0)=y$
$0=f(y)$
$0=\int_{2}^{y} e^{u^2}du$
The value of integration will be zero if $y=2$ so $f^{-1} (0)$ exists and its value is 2
d) What does $(f^{-1})' (0) $ mean? The derivative of inverse function or the inverse of the derivative function?
I'm assuming, for our purposes, that we need $a \ge 2$. If it's not, some of the later questions seem a bit strange.
a) You could start with what you have, noting that
$$ \int_2^{a+h} e^{u^2}\; du - \int_2^{a} e^{u^2}\; du = \int_a^{a+h} e^{u^2}\; du. $$
Perhaps you can do something from there. However, it's probably simpler to use the Fundamental Theorem of Calculus, which says that
$$ \frac{d}{dx} \int_a^x f(t) \; dt = f(x).$$
b) You are correct in that your function is one-to-one. An important point that your argument should include, though, is that your integrand is always positive -- so you can't start subtracting areas for certain values of $a$.
A more direct argument: If, from the above step, you get an explicit formula for $f(a)$, you can check directly whether $f$ is one-to-one. In order for a function $g$ to be one-to-one, we must have $g(x) = g(y)$ mean that $x = y$. For your purposes, you can set $2 \ge a < b$ and show that $f(a) < f(b)$ or $f(a) > f(b)$.
c) This looks good already!
d) Your response in the comments is good -- for part a, though, not d :). That is, you found $f^\prime$, not $f^{-1} = g$ (in exactly the way I suggested above). Differentiation and integration are inverse operations, in a sense. But those are operations on functions. Inverse functions do different things to numbers.
There's a formula you can use to find the derivative of an inverse function; if you don't recall it, you can recover it by noting that
$$ f(f^{-1}(x)) = x$$
and differentiating both sides. Because of the chain rule, you'll be able to solve for $(f^{-1})^\prime(x)$.
In this case, it isn't difficult to find $(f^{-1})^\prime$ directly by finding $f^{-1}$ and then differentiating, but maybe it's good practice the other way as well (and not especially difficult).
As for the assumption that $a \ge 2$: We could still do everything if we let $a$ be a little bit smaller. (How much?) But if we go too low, we have to say no to some of those later questions: The items being asked for do not exist. If you think that's the intention of the question, then you can adjust accordingly.