An Inverse Inequality

Calculus Level 2

Let $f(x) = x^3 + ax^2 + bx + c$ , where $a, b$ , and $c$ are real numbers . In order for $f(x)$ to be invertible, $a$ and $b$ must be related as: $\dfrac{a^m}{b^n} \leq p$ , where $m, n$ , and $p$ are also real numbers.

Find the mimimum value of $m + n + p$ .

The answer is 6.

1 solution

M M
Jun 16, 2016

The Calculus approach:

Note that in order for a cubic function to be invertible, it must either be monotonically increasing or monotonically decreasing. It can have no local minima or maxima, only a single inflection point.This means that $f'(x) \geq 0$ or $f'(x) \leq 0$ for $x \in \mathbb{R}$

Note that $f'(x) = 3x^2 + 2ax + b$ , which is a quadratic. This quadratic can have at most one root, which means its discriminant must be less than or equal to 0.

Therefore,

$(2a)^2 - 4 (3) (b) \leq 0$

$4a^2 - 4 (3) (b) \leq 0$

$a^2 - 3b \leq 0$

$a^2 \leq 3b$

$\frac{a^2}{b} \leq 3$

2 + 1 + 3 = 6.

A purely algebraic approach is forthcoming but more of a pain. The approach I personally would take is finding the vertex (h,k) in terms of a, b, and c. Then the function is invertible iff f(x)=k has only one real solution. This could be determined using the cubic discriminant.

did the same way +1

Pawan pal - 4 years, 11 months ago

An Inverse Inequality

The answer is 6.

1 solution

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