The function $f \colon \mathbb R \to \mathbb R$ is a degree $5$ polynomial which returns $0$ for $x=-3$ and $x=-5$

Question

The function $f \colon \mathbb R \to \mathbb R$ is a degree $5$ polynomial which returns $0$ for $x=-3$ and $x=-5$, and it is known that $f'(-1)=f'(1)=-1$.

What can we say about its zeroes? What can we say about the zeros of $f'$ and $f''$? What can we say if we know that the polynomial has no multiple zeroes? What can we say about its zeroes if we also know that $f(1)=f(-1)=2$?

I've been trying to solve this problem. Since $f'(-1)=f'(1)=-1$ the derivative $f'$ is an odd function, and that's the only thing I could find.

Answer 1

Hint: Solve $$P(-3)=a(-3)^5+b(-3)^4+c(-3)^2-3d+e=0$$ $$P(-5)=a(-5)^5+b(-5)^4+c(-5)^2-5d+e=0$$ $$5a-4b-2c+d=-1$$ $$5a+4b+2c+d=-1$$ if $$P(x)=ax^5+bx^4+cx^3+dx^2+ex+d$$

Answer 2

Here's how you can reconstruct the polynomial given your information: we know that $f$ has the form $$ f(x) = a x^5 + b x^4 + c x^3 + d x^2 + ex + f. $$ Therefore, its derivative is given by $$ f'(x) = 5 a x^4 + 4 b x^3 + 3 c x^2 + 2 d x + e. $$ The information given yields $$ f(-3) = -a 3^5 + b 3^4 - c 3^3 + d 3^2 -3e + f = -243a + 81b - 27c + 9d - 3e + f \overset{!}{=} 0 $$ and $$ f(-5) = -a 5^5 + b 5^4 - c 5^3 + d 5^2 -5e + f = -3125a + 625b - 125c + 25d - 5e + f \overset{!}{=} 0 $$ as well as $$ f'(-1) = 5 a - 4 b + 3c - 2d + e \overset{!}{=} - 1 $$ and $$ f'(1) = 5 a + 4 b + 3c + 2d + e \overset{!}{=} - 1. $$ If we also know $f(1) = f(-1) = 2$, we can derive similar equations $$ a + b + c + d + e +f = 2 \qquad \text{and} \qquad 2 = -a + b - c + d - e + f. $$ We now have six linearly independent equations for the six coefficients, so the linear system $$ \begin{pmatrix} -243 & 81 & -27 & 9 & - 3 & 1 \\ -3125 & 625 & -125 & 25 & - 5 & 1 \\ 5 & - 4 & 3 & - 2 & 1 & 0 \\ 5 & 4 & 3 & 2 & 1 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ - 1 & 1 & -1 & 1 & - 1 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \\ d \\ e \\ f \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ - 1 \\ - 1 \\ 2 \\ 2 \end{pmatrix}, $$ has the unique solution $$ a = -\frac{1}{18}, \qquad b = -\frac{37}{96}, \qquad c = -\frac{7}{18}, \qquad d = \frac{37}{48}, \qquad e = \frac{4}{9}, \qquad f = \frac{155}{96}, $$ so your polynomial is $$ f(x) = \frac{1}{288} \left(-16 x^5 - 111 x^4 - 112 x^3 + 222 x^2 + 128 x + 465\right), $$ which has another real root $\approx 1.5480$ and two complex roots $\approx - 0.24276 \pm 1.09209 i$.

The zeros of $f'$ are approximately $-4.29$, $-1.81$, $-0.25$ and $0.81$ and the roots of $f''$ are approximately $-3.43$, $-1.01$ and $0.37$.

Note. Another way to reconstruct $f$ would have been to write $f(x) = (x + 3)(x + 5)( A x^3 + B x^2 + C x + D)$, so only four coefficients $A, B, C, D \in \mathbb R$ have to be determined.

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If we only consider the first four equations, any $a, b \in \mathbb R$ with $$ \begin{cases} 4 6 c = - 718 a + 128 b + 1 \\ d = - 2 b \\ 4 6 e = 2039 a - 384 b - 49 \\ 23 f = 7680 a + 855 b - 60 \end{cases} $$ are solutions.

Update: Insights without lengthy computations

Note that the only simple equation, $d = - 2 b$ can also be obtained from the fact that $f'(-1) = f'(1)$, because this implies $\int_{-1}^{1} f''(x) = f'(1) - f'(-1) = 0$, which we can rewrite as $8 b + 4 d = 0$.

Furthermore, $f'(1) = f'(-1) \ne 0$ implies that $f'$ has a zero between $-1$ and $1$. (Another possibility would be for $f'$ to be constant, but then $f(x) = e x + f$ could not have two zeros.)

Since the degree of $f$ is five, it has two real roots and and complex roots always occur in complex conjugate pairs, there are only two possibility for the roots:

1. all five roots are real,
2. there are three real roots and two complex roots.