Given the function $\displaystyle f(x)=\frac{A}{x^2-(\alpha +\beta)x+\alpha\beta}$, state the equations of all the asymptotes and find the turning points, stating the domain.
Please show complete working out. I don't know how to find out the above.
Given the function $\displaystyle f(x)=\frac{A}{x^2-(\alpha +\beta)x+\alpha\beta}$, state the equations of all the asymptotes and find the turning points, stating the domain.
Please show complete working out. I don't know how to find out the above.
Ok,
notice that
$\displaystyle x^2-(\alpha+\beta)x+\alpha\;\beta=(x-\alpha)(x-\beta)$
This shows you the vertical asymptotes, as the values of x that cause the denominator to be zero
are evident.
The horizontal asymptote is discovered by letting x approach infinity.
Your previous method of partial fractions is useful for calculating the derivative of quotients.
Alternatively, use the Quotient Rule.
However, in this case, it's convenient to use a "negative exponent" instead.
$\displaystyle f(x)=\displaystyle\frac{A}{x^2-(\alpha+\beta)x+\alpha\;\beta}=A\left[x^2-(\alpha+\beta)x+\alpha\;\beta\right]^{-1}$
Using the Chain Rule
$\displaystyle u=x^2-(\alpha+\beta)x+\alpha\;\beta$
$\displaystyle \displaystyle\frac{d}{dx}Au^{-1}=A\frac{du}{dx}\;\frac{d}{du}u^{-1}$
$\displaystyle =\displaystyle\ A\left[2x-(\alpha+\beta)\right]\left(-\left[u^{-2}\right]\right)=\frac{-A\left[2x-(\alpha+\beta)\right]}{\left[x^2-(\alpha+\beta)x+\alpha\;\beta\right]^2}$
This is zero when the numerator is zero, if we do not have an $\displaystyle (x-\alpha)$ or $\displaystyle (x-\beta)$ factor there.
(all the work calculating the denominator is often a waste of time, but not always).
Since $\displaystyle A\ne\ 0$
$\displaystyle \displaystyle\ 2x-(\alpha+\beta)=0\Rightarrow\ x=\frac{\alpha+\beta}{2}$
There is a turning point at that x, provided $\displaystyle \alpha\ne\ \beta$,
otherwise, we'd have calculated a turning point at the asymptote, which is not possible.
Then place that x into $\displaystyle f(x)$ to find the vertical co-ordinate of the turning point.
Alright, alright, makes sense. I don't really pretend to be good at all this maths stuff.
I'm also not getting the correct corresponding y value when I algebraically substitute the x value of the turning point in f(x). Would anyone be able to supply the rest of the working out to see where I went wrong?
$\displaystyle f(\frac{\alpha+\beta}{2}) = A[(\frac{\alpha+\beta}{2})^2 -(\alpha+\beta)+\alpha\beta]^{-1}
=A[\frac{\alpha^2 +\beta^2}{4} - \frac{4(\alpha+\beta)}{4} +\frac{4\alpha\beta}{4}]^{-1}$
$\displaystyle =A[\frac{\alpha^2+\beta^2-4\alpha+4\beta+4\alpha\beta}{4}]^{-1}$
Where have I gone wrong?