# Turning point

• Jul 24th 2010, 12:27 AM
kingman
Turning point
Dear Sir,
I would be most grateful if some friends can help to solve the attached question.
Thanks
• Jul 24th 2010, 06:32 AM
Soroban
Hello, kingman!

I'm coming up with a strange result . . .

Quote:

A curve has the equation: . $y \:=\:\dfrac{2x^2+px+2}{x+1}$ where $p$ is a constant.

By considering $\dfrac{dy}{dx}$ .
. . . I don't see how
. . find the range of $p$ such that the equation: . $\dfrac{2x^2+px+2}{x+1} \:=\:k$
. . has exactly 2 real roots for all real values of $k.$

The equation is: . $2x^2 + px + 2 \:=\:k(x+1)$

. . which simplifies to: . $2x^2 + (p-k)x + (2-k) \:=\:0$

A quadratic has two real roots if its discriminant is positive.

The discriminant is: . $b^2 - 4ac \;=\;(p-k)^2 - (4)(2)(2-k)$

. . And we want: . $p^2 - 2kp + k^2 + 8x - 16 \;>\;0$ .[1]

This is an up-opening parabola.
. . It will be positive if its vertex is above the $p$-axis.

Its vertex is found at: . $p \:=\:\dfrac{\text{-}b}{2a} \:=\:\dfrac{\text{-}(\text{-}2k)}{2(1)} \:=\:k$

. . Hence, the vertex is: . $(k,\;8k-16)$

The parabola is positive if: . $8k-16 \:>\:0 \quad\Rightarrow\quad k \:>\:2$

We have a constaint on $k.$

Yet we must find values of $p$ which satisfy [1] for all values of $k.$

Is there a flaw in my reasoning?

• Jul 24th 2010, 07:24 AM
kingman
If eq 1 is written in powers of k instead of p we can get the answer but I still wonder why.But question required one to start from dy/dy with the hint that if dy/dy =0 has no turning point than the answer follows and again I wonder why this is so.
Thanks