Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The following two functions

$$f_1(x) = \frac{(3-x)(1-x)}{3}$$

and

$$f_2(y) = \frac{(1-y^2)}{(1+y) (2(1+y) - y(1-y)^2) }$$

are given.

Find range of values of $x$ and $y$ so that

$$f_1(x) < f_2(y)$$

share|improve this question
2  
$x=1$, $y=0{}{}$ –  martini Nov 15 '12 at 13:12

1 Answer 1

up vote 1 down vote accepted

We have $$ 2(1 + y) - y(1-y)^2 = 2 + 2y - y + 2y^2 - y^3 = -y^3 + 2y^2 + y + 2$$ and hence $$ f_2(y) = \frac{1-y}{-y^3 + 2y^2 + y + 2} $$ Now let $y$ be given, we will find the $x \in \mathbb R$ with $f_1(x) < f_2(y)$, we have \begin{align*} f_1(x) &< f_2(y)\\ \iff x^2 - 4x + 3 &< 3f_2(y)\\ \iff (x-2)^2 - 1 &< 3f_2(y)\\ \iff (x-2)^2 &< 3f_2(y) + 1\\ \end{align*} We have \begin{align*} 3f_2(y) + 1 &= \frac{-y^3+2y^2-2y + 5}{-y^3 + 2y^2 + y + 2}\\ &= \frac{y^3 -2y^2 + 2y - 5}{y^3 - 2y^2 - y -2} \end{align*} As we want $(x-2)^2 < 3f_2(y) + 1$, we want this to be positive, Wolfram|Alpha tells us (I'm sure, one can do this by hand using Cardano's formulas to find the points where numerator and denominator change sign) that this is the case for $\def\ytwo{\frac13\left(2 + \sqrt[3]{44-3\sqrt{177}}+\sqrt[3]{44+3\sqrt{177}}\right)}\def\yone{\frac 13\left(2 - 2 \sqrt[3]{\frac 2{155 + 3\sqrt{1473}}}+\sqrt[3]{\frac12\left(155 + 3\sqrt{1473}\right)}\right)}$ \[ y \not\in \left[\yone, \ytwo\right] \] Then we can continue $$ (x-2)^2 < 3f_2(y) + 1 \iff |x-2| < \sqrt{3f_2(y) + 1} $$ that gives $\def\ftwoy{\frac{y^3 -2y^2 + 2y - 5}{y^3 - 2y^2 - y -2}}$ $$ x \in \left(2-\sqrt{\ftwoy}, 2+\sqrt{\ftwoy}\right). $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.