Bounded and cubic

Algebra Level 4

Consider a general cubic polynomial function $f(x) = ax^3 + bx^2 + cx + d$ with real constants $a>0 , b>0, c\ne 0, d \ne 0$ such that for all $|x| \leq 1$ , the inequality $|f(x) | \leq 1$ is fulfilled.

Find $\max \left(a + b + |c| + |d| \right)$ .

Notation: $| \cdot |$ denotes the absolute value function .

The answer is 7.

1 solution

Archit Tripathi
Dec 7, 2016

$\frac{1}{c}, \frac{1}{d} > 0$

$f(1)\leq 1$

$\frac{1}{d} < 0, \frac{1}{c} > 0$

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} - \frac{1}{d} = f(1) - 2f(0)\leq 3$

$\frac{1}{c} < 0, \frac{1}{d} > 0$

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} = \frac{4}{3}f(1) - \frac{1}{3}f(-1) - \frac{8}{3}f(\frac{1}{2}) + \frac{8}{3}f(-\frac{1}{2})$ $\leq \frac{4}{3} + \frac{1}{3} + \frac{8}{3} + \frac{8}{3} = \boxed{7}$ .

$\frac{1}{c}, \frac{1}{d} < 0$

$\frac{1}{a} + \frac{1}{b} - \frac{1}{c} - \frac{1}{d} = \frac{5}{3}f(1) - 4f(\frac{1}{2}) + \frac{4}{3}f(-\frac{1}{2})$ $\leq \frac{5}{3} + 4 + \frac{4}{3} = \boxed{7}$ .

Very interesting question :) We made edits to simplify the phrasing in the problem. Can you review it for accuracy and then update your solution accordingly?

Also note that you've only found a necessary condition. But is it sufficient? IE in finding a maximum, you have demonstrated that 7 is an upper bound. However, is this the lowest upper bound? Can it be achieved for some value of a, b, c, d? If so, we should state it (and ensure that the conditions in the problem are satisfied).

Calvin Lin Staff - 4 years, 6 months ago

K, it is indeed a lowest upper bound.

Equality is achieved with $f(x) = 4x^3 - 3x$ , which satisfies $f(1) = 1, f( -1) = -1 , f(1/2) = -1/2, f(-1/2) = 1$ as given in your second case.

Calvin Lin Staff - 4 years, 6 months ago

Firstly, there appears to be a typo in the sixth line

Secondly, there seems to be lacking rigour in your proof as u seek to establish an upper bound without proving its attainability.

For example, showing $x^2$ exceeds -1 doesn't establish -1 as it's minima

Note: I may be mistaken. Please tell me why if so

Rohith M.Athreya - 4 years, 6 months ago

It took me a lot of time to finally understand your presentation. Eg. I didn't realise it's actually 4 case studies. I believe adding words would be easier for people to follow!

Christopher Boo - 4 years, 6 months ago

Bounded and cubic

The answer is 7.

1 solution

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