Given that $F(x)=\int^x_0e^{t^2}dt+1$, which of these is true? Let $F(x)=\int^x_0e^{t^2}dt+1$. Then: 

a) $F(x)$ is convex in $(-\infty , \infty )$. 
b) The equation $F(x)=0$ has two solutions. 
c) $F(x)$ is invertible and $(F^{-1})'(1)=1$. 
d) $F(1)=5$. 
e) $F(x)=F(-x)$ for every $x$.

My attempt:
a) $F'(x)=e^{x^2}$, $F''(x)=2xe^{x^2}<0$ for $x<0$ so it's not true. 
b) I don't know how to disprove this one. 
c) I know that $(F^{-1})'(F(x))=\frac {1}{F'(x)}$, I hesitated a lot here, since if I want to get $(F^{-1})'(1)$, I need to substitute $x=0$, but do I substitute also $x=0$ in $F'(x)$ or $x=1$? (this was  critical to know for this question). 
d) I got no idea just pure intuition that the area at $x=1$ of $e^{x^2}$ would be less than $5$. 
e) That's not true, it changes the integral, and will somewhere affect the $e$ being with $(-)$ in the solution.
The answer was c). I would really appreciate any help in disproving the wrong answers and any tips regarding my confusion about c).
 A: (b) is wrong because $F'(x) = e^{x^2} > 0$, so that $F$ is strictly increasing.
For (c) one can use that $F'(x) = e^{x^2} \ge 1$ implies that the range of $F$ is $\Bbb R$ and $F$ is invertible. Then
$$
 (F^{-1})'(1) = \frac{1}{F'(F^{-1}(1))}=  \frac{1}{F'(0)} = \frac{1}{e^{0^2}} = 1 \, .
$$
A: The key to quickly answer these multiple choice questions is to identify the largest commonality, which in this case is the sign of the first derivative.
The fact that $F'(x)=e^{x^2}$ means $F$ is strictly increasing. This immediately eliminates

*

*b) as $F$ crosses the horizontal axis at most once;

*e) as no even function is strictly increasing on the entire domain, and

*d) as $\int_0^1e^{t^2}\,dt<(1-0)\times e^{1^2}$.

We can eliminate a) by differentiating again to check the sign, which you have correctly done.
With c), we know that $F$ is invertible as it is continuous and its range is the set of real numbers. The inverse derivative formula which is derived from using the chain rule on $F^{-1}(F(x))=x$, can be used to evaluate $(F^{-1})'(1)=[F'(F^{-1}(1))]^{-1}=F'(0)^{-1}$ as $F$ is bijective.
Thus c) is the only correct answer.
