Re: Elementary Calculus
Hello, atomos!
I think I've solved part (a) . . .
\(\displaystyle y\:=\:\frac{x^2\,-\,1}{x^2\,+\,2x\,-\,1}\,\) is curve \(\displaystyle C.\)
(a) Show(without drawing) that line \(\displaystyle y\,=\,p\;(p\,\neq\,1)\,\) will always cross the curve \(\displaystyle C\) in two points, \(\displaystyle A\) and \(\displaystyle B.\)
(b) What should be the value of \(\displaystyle p\) so that points \(\displaystyle A\) and \(\displaystyle B\) are always in the range of \(\displaystyle x\,>\,0\) ?
Where are the intersections? . . . Set the functions equal to each other and solve.
We have: \(\displaystyle \;\frac{x^2\,-\,1}{x^2\,+\,2x\,-\,1}\:=\
\;\;\Rightarrow\;\;x^2\,-\,1\:=\
x^2\,+\,2px\,-\,p\)
This is a quadratic: \(\displaystyle \,(p-1)x^2\,+\,2px\,-\,(p-1)\:=\:0\)
Quadratic Formula: \(\displaystyle \,x\;=\;\frac{-2p\,\pm\,\sqrt{(2p)^2\,+\,4(p-1)(p-1)}}{2(p-1)}\;=\;\frac{-p\.\pm\,\sqrt{2p^2\,-\,2p\,+\,1}}{p-1}\)
There will be two values of \(\displaystyle x\) (two intersections) if the discriminant is positive.
\(\displaystyle \;\;\)That is: \(\displaystyle \,2p^2\,-\,2p\,+\,1\:>\:0\)
When is this true?
We have a parabola: \(\displaystyle y\:=\:2p^2\,-\,2p\,+\,1\) which opens upward.
\(\displaystyle \;\;\)Its minimum is where \(\displaystyle y'\,=\,0:\;4p\,-\,2\:=\:0\;\;\Rightarrow\;\;p\,=\,\frac{1}{2}\)
\(\displaystyle \;\;\)Then: \(\displaystyle y\:=\:2\left(\frac{1}{2}\right)^2 \,-\,2\left(\frac{1}{2}\right)\,+\,1\:=\;\frac{1}{2}\)
Hence, the vertex (lowest point) is: \(\displaystyle \,\left(\frac{1}{2},\,\frac{1}{2}\right)\)
The entire parabola is "positive".
Therefore, the discriminant is always positive
\(\displaystyle \;\;\)and there
will be two points of intersection, \(\displaystyle A\) and \(\displaystyle B.\)
