Prove that every tangent line to the function $f(x)$ only intersects the function once.
$$f(x) = \int_{0}^{x^2}e^{t^3}dt, x \in(0,\infty)$$
My Attempt:
Lemma 1: $f'(x)$ is an increasing function.
By the fundamental theorem of calculus we get $f'(x) = 2xe^{x^6}$. Assume $x_1, x_2 \in (0, \infty)$ such that $x_1 < x_2$. Then:
$$2x_1 < 2x_2$$ $$2x_1e^{x_1^6} < 2x_2e^{x_2^6} \rightarrow f'(x_1) < f'(x_2)$$
This lemma allows us to conclude that every slope of $f(x)$ is unique. Assume that there is an intersection point for some slope. Let the slope be at $(x_1, y_1)$ and the intersection point be $(a,b)$. By the MVT there is a $f'(c) = f'(x_1)$ where $c \in (x_1, a)$. This contradicts lemma 1. QED
Is this strategy correct?
The strategy is good. There is a small, somewhat inconsequential error though: the derivative of $f$ is $2xe^{x^6}$ not $2xe^{x^6} + C$. Remember, you're not integrating here, but differentiating a function (defined by a definite integral, which wouldn't need a constant of integration anyway).
What you're essentially doing here is showing that $f$ is a strictly convex function; if a differentiable function $f$ has a strictly increasing derivative, then $f$ is strictly convex. Tangents to convex functions lie on or below the graph of the function. In the case of strictly convex functions, the tangent only touches the function once.