Let $n$ be a positive integer. For what values of $n$ does every tangent line to the graph $y=x^n$ intersect the graph exactly once?
I said we have $\frac{dy}{dx} = nx^{n-1}$ and at the point $(x_0,y_0)$, we have $y = nx_0^{n-1}x+(y_0-nx_0^n)$. Thus our equation is $nx_0^{n-1}x+(y_0-nx_0^n) = x^n$ which must have exactly one intersection. I am not sure how to show this, though.
This was my answer before I noticed the restriction to positive integer $n$. This answer can be restricted to that situation.
If $n$ is a positive even integer, $y=x^n$ is concave up, and tangent lines only touch the curve at the point of tangency.
If $n$ is a negative odd integer, $y=x^n$ has two disjoint arms. The right arm is concave up, and tangent lines to the right arm only touch the curve at the point of tangency. Such lines cannot touch the other arm because they never enter the third quadrant. Likewise for tangent lines to the left arm.
If $n$ is a negative even integer, it's clear that the tangent line at $(1,1)$ will also cross the left arm, since it has negative slope and the left arm climbs to $\infty$.
If $n$ is a positive odd integer, then the tangent line at $(1,1)$ will also cross the curve in the third quadrant because the curve is concave down in that quadrant, and the line is below the curve at the $y$-axis.
If $n$ is $0$, of course tangent lines are the curve itself.
If $n$ is not an integer, then $x^n$ is only defined for $x\geq0$ and the curve is either concave up or concave down in this region (but doesn't toggle). So tangent lines only touch the curve at the point of tangency.
In summary, the collection of such $n$ is $$\left(1-2\mathbb{N}\right)\cup(\mathbb{R}\setminus\mathbb{N})\cup(2\mathbb{N})$$ (where I'm using the convention $\mathbb{N}=\{1,2,\ldots\}$)
Hint: If n are even, is clear that works. If n are odd, note that the function is increasing while the derivate not because have a zero and is non negative.
