Find the largest and smallest values of the functions
a) \(f_1 (x) = a \cos x + b \sin x\); b) \(f_2 (x) = a \cos^2x + b \cos x \sin x + c \sin^2x\).
Prove that the tangent to the graph of the function \(f (x)\), constructed at coordinates \((x_0, f (x_0))\) intersects the \(Ox\) axis at the coordinate: \(x_0 -\frac{f(x_0)}{f'(x_0)}\).
The Newton method (see Problem 61328) does not always allow us to approach the root of the equation \(f(x) = 0\). Find the initial condition \(x_0\) for the polynomial \(f(x) = x (x - 1)(x + 1)\) such that \(f(x_0) \neq x_0\) and \(x_2 = x_0\).
Prove the inequality: \[\frac{(b_1 + \dots b_n)^{b_1 + \dots b_n}}{(a_1 + \dots a_n)^{b_1 + \dots + b_n}}\leq \left(\frac{b_1}{a_1}\right)^{b_1}\dots \left( \frac{b_n}{a_n}\right)^{b_n}\] where all variables are considered positive.
Prove that if the function \(f (x)\) is convex upwards on the line \([a, b]\), then for any distinct points \(x_1, x_2\) in \([a; b]\) and for any positive \(\alpha_{1}, \alpha_{2}\) such that \(\alpha_{1} + \alpha_ {2} = 1\) the following inequality holds: \(f(\alpha_1 x_1 + \alpha_2 x_2 ) > \alpha_1 f (x_1) + \alpha_2 f(x_2)\).
Inequality of Jensen. Prove that if the function \(f (x)\) is convex upward on \([a, b]\), then for any distinct points \(x_1, x_2, \dots , x_n\) (\(n \geq 2\)) from \([a; b]\) and any positive \(\alpha_{1}, \alpha_{2}, \dots , \alpha_{n}\) such that \(\alpha_ {1} + \alpha_{2} + \dots + \alpha_{n} = 1\), the following inequality holds: \(f (\alpha_{1} x_1 + \dots + \alpha_{n} x_n) > \alpha_{1} f (x_1) + \dots + \alpha_{n} f (x_n)\).
Prove that the polynomial \(P (x)\) is divisible by its derivative if and only if \(P (x)\) has the form \(P(x) = a_n(x - x_0)^n\).
Prove that for \(n > 0\) the polynomial \[P (x) = n^2x^{n + 2} - (2n^2 + 2n - 1) x^{n + 1} + (n + 1)^2x^n - x - 1\] is divisible by \((x - 1)^3\).
Prove that for \(n> 0\) the polynomial \(x^{2n + 1} - (2n + 1)x^{n + 1} + (2n + 1)x^n - 1\) is divisible by \((x - 1)^3\).
Prove that the polynomial \(P (x) = a_0 + a_1x + \dots + a_nx^n\) has a number \(-1\) which is a root of multiplicity \(m + 1\) if and only if the following conditions are satisfied: \[\begin{aligned} a_0 - a_1 + a_2 - a_3 + \dots + (-1)^{n}a_n &= 0,\\ - a_1 + 2a_2 - 3a_3 + \dots + (-1)^{n}na_n &= 0,\\ \dots \\ - a_1 + 2^{m}a_2 - 3^{m}a_3 + \dots + (-1)^{n}n^{m}a_n &= 0. \end{aligned}\]