Polynomials (Table of contents)
# Applications of Calculus

**Problem 17**
**Theorem 6.1 **
**Problem 18**
**Problem 19**
**Theorem 6.2 (Rolle’s Theorem) **
**Corollary **

The derivative of a polynomial \(P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\) is given by \[P^{\prime}(x)=na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+\cdots+a_1.\] The inverse operation, the indefinite integral, is given by \[\int P(x)dx=\frac{a_n}{n+1}x^{n+1}+\frac{a_{n-1}}nx^n+\cdots+ a_0x+C.\] If the polynomial \(P\) is not given by its coefficients but rather by its canonical factorization, as \(P(x)=(x-x_1)^{k_1}\cdots(x-x_n)^{k_n}\), a more suitable expression for the derivative is obtained by using the logarithmic derivative rule or product rule: \[P^{\prime}(x)=P(x)\left(\frac{k_1}{x-x_1}+\cdots+ \frac{k_n}{x-x_n}\right).\] A similar formula can be obtained for the second derivative.

Suppose that real numbers \(0=x_0 < x_1 < \cdots < x_n < x_{n+1}=1\) satisfy \[\sum_{j=0,\,j\neq i}^{n+1}\frac{1}{x_i-x_j}=0\quad\mbox{for } i=1,2,\dots,n.\quad\quad\quad\quad\quad(1)\] Prove that \(x_{n+1-i}=1-x_i\) for \(i=1,2,\dots,n\).

What makes derivatives of polynomials especially suitable is their property of preserving multiple zeros.

If \((x-\alpha)^k\mid P(x)\), then \((x-\alpha)^{k-1}\mid P^{\prime}(x)\).

Determine a real polynomial \(P(x)\) of degree at most 5 which leaves remainders \(-1\) and 1 upon division by \((x-1)^3\) and \((x+1)^3\), respectively.

For polynomials \(P(x)\) and \(Q(x)\) and an arbitrary \(k\in\mathbb{C}\), denote \[P_k=\{z\in\mathbb{C}\mid P(z)=k\}\quad\mbox{and}\quad Q_k=\{z\in\mathbb{C}\mid Q(z)=k\}.\] Prove that \(P_0=Q_0\) and \(P_1=Q_1\) imply that \(P(x)=Q(x)\).

Even if \(P\) has no multiple zeros, certain relations between zeros of \(P\) and \(P^{\prime}\) still hold. For example, the following statement holds for all differentiable functions.

Between every two zeros of a polynomial \(P(x)\) there is a zero of \(P^{\prime}(x)\).

If all zeros of \(P(x)\) are real, then so are all zeros of \(P^{\prime}(x)\).