1. Proof

Prove that $\frac{2x^2 + 5x + 2}{3x^2 + 4x + 1}$ can have any real value $\forall x\in R$.

I assumed the expression to be equal to $y$. And on simplification I got a quadratic in $x$. Since $x\in R$ the discriminant $D\geqslant 0$ . I dont know where to go from here..... Please help.

2. This function is the ratio of two continuous polynomials, so will also be continuous except where the denominator is zero.

\displaystyle \begin{align*}3x^2 + 4x + 1 &= 0\\ 3x^2 + 3x + x + 1&= 0\\ 3x(x + 1) + 1(x + 1) &= 0\\ (x + 1)(3x + 1) &= 0\end{align*}

So the function is discontinuous at $\displaystyle x = -1$ and $\displaystyle x = -\frac{1}{3}$

Your statement that this function can be evaluated for all real $\displaystyle x$ is wrong.

3. Originally Posted by mathlover14
Prove that $\frac{2x^2 + 5x + 2}{3x^2 + 4x + 1}$ can have any real value $\forall x\in R$.
You want to try to state the problem more clearly. What you've written doesn't quite make sense. You'd be surprised what clear phrasing can do for your ability to solve a problem. The correct phrasing of the problem is:

$\forall y \in \mathbb{R}: \exists x \in \mathbb{R}: \frac{2x^2 + 5x + 2}{3x^2 + 4x + 1}=y$

Then the solution to your problem is very straightforward. Start with the phrase: "Choose y" or "Let y be an arbitrary real number." Then the job of your proof is to FIND (construct, prove the existence of, etc) some x such that

$\frac{2x^2 + 5x + 2}{3x^2 + 4x + 1}=y$

Once you've proven the result for an "arbitrarily chosen" y, you're done. That is, once you find that x, you've got nothing more to do! Saying "I assumed the expression to be equal to y" isn't exactly what you mean. And by the way, if it helps, Let y be, for example, 2. Then ask yourself how you would solve the above equation. Then generalize the result: show that equation is solvable not only if y is 2, but if it is anything. See?