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.