Spivak Calculus 4th Ed. Chapter 11 Problem 27 I am having trouble understanding what the following problem is getting at. I've included my attempted answers below each part.
Chapter 11 Problem 27
(a) Suppose the polynomial function $f(x) = x^n + a_{n-1}x^{n-1} + .. + a_0$ has exactly $k$ critical points and that $f''(x) \neq 0$ for all critical points $x$. Show that $n - k$ is odd.
Since $f''(x) \neq 0$ all critical points are local minima or maxima. The sign of the derivative must switch at and only at each of these points since $f'$ is a polynomial. The sign of the derivative must also match the beginning and ending behavior of $f$. For example, $f'$ starts negative and ends positive if $n$ is even. So for even $n$, $k$ must be odd and for odd $n$, $k$ must be even.
(b) For each $n$, show that if $n - k$ is odd, then there is a polynomial function $f$ of degree $n$ with $k$ critical points, at each of which $f''$ is nonzero.
Let $f'(x) = (x - 1)(x - 2)...(x - k)(x^{n - k - 1} + 1)$ and integrate. $n - k - 1$ is even so $f$ has exactly $k$ critical points. Since each critical point $x$ is a single root of $f'$, $f''(x) \neq 0$ (by product rule).
(c) Suppose that the polynomial function $f(x) = x^n + a_{n-1}x^{n-1} + .. + a_0$ has $k_1$ local maximum points and $k_2$ local minimum points. Show that $k_2 = k_1 + 1$ if $n$ is even, and $k_2 = k_1$ if $n$ is odd.
Seems like the argument in (a) also works here. Local maxima and minima must alternate and the restrictions on the end behavior enforce the given relationships.
(d) Let $n, k_1, k_2$ be three integers with $k_2 = k_1 + 1$ if $n$ is even, and $k_2 = k_1$ if $n$ is odd, and $k_1 + k_2 < n$. Show that there is a polynomial function $f$ of degree $n$, with $k_1$ local maximum points and $k_2$ local minimum points. Hint: Pick $a_1 < a_2 < ... < a_{k_1 + k_2}$ and try $f'(x) = \prod_{i = 1}^{k_1 + k_2}(x - a_i) \cdot (1 + x^2)^{l}$ for an appropriate number $l$.
Similarly, seems like part (b) works here. All the critical points $x$ are local minima or maxima since $f''(x) \neq 0$ and the sign switching argument implies that there must be the correct number of each. However, this does not use the hint. Is this still correct? What is he getting at in the hint?

If the polynomial $f'$ has $k$ real roots, and $f''$ does not vanish in any of them, then all its $k$ roots are simple. Hence, $f'$ possesses exactly $k$ real roots (counting multiplicities), and all the remaining roots (if any), their number is $n-1-k$, are complex. But a real polynomial has an even number of complex roots.
Thus $n-1-k$ is even and $n-k$ is odd.
