Hoping to get some verification on this proof.
Calculus by Michael Spivak, 3rd Ed. Chapter 11, exercise 58:
58. Use the method of Problem 57 to prove that if $n$ is even and $f(x) = x^n$, then every tangent line to $f$ intersects $f$ only once.
Here's the referenced exercise:
57. (a) Let $y \neq 0$ and let $n$ be even. Prove that $x^n + y^n = (x + y)^n$ only when $x=0$. Hint: If $x_0^n + y^n = (x_0 + y)^n$, apply Rolle's Theorem to $f(x) = x^n + y^n - (x+y)^n$ on $[0,x_0]$.
(b) Prove that if $y \neq 0$ and $n$ is odd, then $x^n + y^n = (x+y)^n$ only if $x = 0$ or $x = -y$.
My proof is a little different than the one in the Answer Book, so I was hoping someone could look it over and see if it's ok.
58. Solution
We have $f(x) = x^n$ (where $n$ is even).
Let's assume the tangent line to $f(x)$ at some point $(a,a^n)$, intersects $f(x)$ at another point $(b,b^n), b\neq a$.
This tangent line has slope $$f^\prime(a) = na^{n-1}$$
$f(x)$ is differentiable on $[a,b]$ so the Mean Value Theorem tells us there exists some point $c$ in the interval $(a,b)$ such that $$f'(c) = \frac{f(b)-f(a)}{b-a}$$ but this is just $f'(a)$.
In other words, we have
$$f'(c) = f'(a)$$ $$nc^{n-1} = na^{n-1}$$ $$c^{n-1} = a^{n-1}$$
Because $n-1$ is odd, this is only possible if $c = a$, a contradiction. So our assumption that $b$ exists such that $(b,b^n)$ is intersected by the tangent of $f$ at $(a,a^n)$ must be wrong.