• Jun 5th 2010, 05:27 PM
may
Show that if x is real ,the expression [(2x-5)(x+1)]/x-1 can take all the real values.
• Jun 5th 2010, 05:29 PM
Prove It
Quote:

Originally Posted by may
Show that if x is real ,the expression [(2x-5)(x+1)]/x-1 can take all the real values.

Is this $\displaystyle \frac{(2x - 5)(x + 1)}{x} - 1$ or $\displaystyle \frac{(2x - 5)(x + 1)}{x - 1}$?

In either case, your statement is false, as you can't divide by $\displaystyle 0$.

So $\displaystyle x \neq 0$ in the first and $\displaystyle x \neq 1$ in the second.
• Jun 5th 2010, 05:34 PM
may
Quote:

Originally Posted by Prove It
Is this $\displaystyle \frac{(2x - 5)(x + 1)}{x} - 1$ or $\displaystyle \frac{(2x - 5)(x + 1)}{x - 1}$?

In either case, your statement is false, as you can't divide by $\displaystyle 0$.

So $\displaystyle x \neq 0$ in the first and $\displaystyle x \neq 1$ in the second.

is the second one $\displaystyle \frac{(2x - 5)(x + 1)}{x - 1}$
• Jun 5th 2010, 05:46 PM
Prove It
Quote:

Originally Posted by may
is the second one $\displaystyle \frac{(2x - 5)(x + 1)}{x - 1}$

Your statement is still false, as $\displaystyle x$ can not take on the value of $\displaystyle 1$.
• Jun 5th 2010, 05:49 PM
skeeter
Quote:

Originally Posted by Prove It
Your statement is still false, as $\displaystyle x$ can not take on the value of $\displaystyle 1$.

I think the problem is referring to the range of the expression rather than the domain.
• Jun 5th 2010, 06:21 PM
Bacterius
Quote:

Originally Posted by may
Show that if x is real ,the expression [(2x-5)(x+1)]/x-1 can take all the real values.

I've never tried this kind of question yet, correct me if I'm wrong.

Let $\displaystyle f(x) = \frac{(2x - 5)(x + 1)}{x - 1}$. If we can establish a full bijection between f$\displaystyle (x)$ and $\displaystyle x$, then $\displaystyle f(x)$ must take all real values. Let's see, say $\displaystyle f(x) = y$ :

$\displaystyle y = \frac{(2x - 5)(x + 1)}{x - 1}$
$\displaystyle x = \frac{1}{4} \left ( - \sqrt{y^2 - 2y + 49} + y + 3 \right )$

The square root is bothersome, as we need to show that the value under is (hopefully) always positive. So :

$\displaystyle y^2 - 2y + 49$

Discriminant is $\displaystyle \Delta = 2^2 - 4 \times 49 = -192 < 0$. Since the quadratic coefficient, $\displaystyle a$, is positive, this quadratic yields only positive values.

Therefore the equation :

$\displaystyle x = \frac{1}{4} \left ( - \sqrt{y^2 - 2y + 49} + y + 3 \right )$

Does accept all values of $\displaystyle y$. And therefore, in :

$\displaystyle y = \frac{(2x - 5)(x + 1)}{x - 1}$

$\displaystyle y$ does take all real values. $\displaystyle f(x)$ does take all real values for $\displaystyle x \in \mathbb{R}$. QED.