Sofia University Entrance Exam

[Math Tutor]

Let's[tex]f(x) = x^3 + ax^2 + bx + c[/tex]
[tex]a, b, c \in R[/tex]
and [tex]|f(x)| \le 2(1 - x^2)[/tex] for every [tex]x \in [-1, 1][/tex].
Find the maximum of the function [tex]|f'(x)|[/tex] in interval [tex]x\in [-1, 1][/tex]

- the problem is from Sofia University(in Bulgaria) entrance exam.

[Guest]

By the condition [tex]|f(x)|\le 2(1-x^2)[/tex] for [tex]x\in [-1,1][/tex] we know [tex]f(x)[/tex] has roots at 1 and -1, so must be of the form [tex]f(x)=(x-1)(x+1)(x+\alpha)=-(1-x^2)(x+\alpha)[/tex] for some [tex]\alpha[/tex]. Furthermore the same condition tells us that [tex]|x+\alpha|\le 2[/tex] for [tex]x\in[-1,1][/tex] which means [tex]\alpha\in[-1,1][/tex].

The derivative [tex]f'(x)=3x^2+2\alpha x-1[/tex] will achieve its maximum absolute value either at the boundary points i.e. when [tex]x=-1[/tex] or [tex]1[/tex] or when [tex]f''(x)=0[/tex] which is at [tex]x=-\alpha/3[/tex]. The corresponding values of [tex]f'(x)[/tex] are [tex]2-2\alpha[/tex], [tex]2+2\alpha[/tex], and [tex]-1-\alpha^2/3[/tex].

Hence the maximum absolute value is 4.

Hope that helps.

R. Baber.