$\int_{3}^{x^2} e^{t^2} +1dt$ minima and inf

Question

$$F(x) = \int_{3}^{x^2} e^{t^2}+1 dt$$

I'm given the following choices:

a. $\inf(F) = -\infty$

b. $\min(F) > 0$

c. $F$ has a lower bound but no minimum

d. $\min(F) < 0$

Here's what I did:

$$F'(x) = F(x^2) \cdot \frac{d}{dx}(x^2) - F(3) \cdot \frac{d}{dx}(3) = 2x(e^{x^4}+1)$$

At this point, since $$e^{x^4} + 1 > 0 \forall x \in\mathbb{R}$$ and $$2x <0 \forall x \in (-\infty,0)$$ the function $F(x)$ is decreasing in the interval $(-\infty,0)$, as the derivative is negative in that interval and only becomes zero in $x=0$.

My answer was therefore (a), but turns out it's wrong. What did I do wrong?

Answer 1

$F$ is decreasing on $(-\infty,0)$ and increasing on $(0,\infty)$ and therefore, the minima is attained at $0$. Also, $$F(0)= \int_3^0e^{x^4}+1dx$$ $$=-\int_0^3e^{x^4}+1dx$$ $$<0$$ So, correct option is (d).