Question

Consider a general case of the function... ?

Consider a general case of the function `f:R\{a,b} -> R, where f(x)=4/((x-a)(x-b))` where a>0 and b>0
a) Find the value of the x-coordinate of the turning point of this graph in terms of a and b
b)Hence, find the value of the local maximum of the function over the interval [a,b] in terms of a and b
c) If b=2, for what values of a will there be only 1 vertical asymptote?

Answer 1

Please see below.

Explanation:

Assuming `a!=b` and `b>a`, converting `f(x)` into partial fractions, we get

`f(x)=4/((x-a)(x-b))=4/(a-b)(1/(x-a)-1/(x-b))`

(i) As turning point appears when `(df)/(dx)=0`, let us find this

`(df)/(dx)=4/(a-b)[-1/(x-a)^2+1/(x-b)^2]`

and `(df)/(dx)=0` when `(x-a)^2-(x-b)^2=0`

i.e. `(a-b)(2x-a-b)=0` i.e. `x=(a+b)/2`

Hence, turning point is at `x=(a+b)/2`

Below is shown the graph if `a=-7` and `b=3`

graph{4/((x+7)(x-3)) [-12.8, 7.2, -5.16, 4.84]}

(ii) At `x=(a+b)/2` it lies between `a` and `b` and

we have `(d^2f)/(dx^2)=4/(a-b)[2/(x-a)^3-2/(x-b)^3]<0`

and we have a local maxima at `x=(a+b)/2`, where

`f(x)=4/(((a+b)/2-a)((a+b)/2-b))=4/((b-a)/2*(a-b)/2)`

= `-16/(b-a)^2`

(iii) Normally `4/((x-a)(x-b))` has two vertical asymptotes, one at `x=a` and other at `x=b`. There will be only one asymptote if `a=b`. Hence if `b=2`, we should have `x=2` and

`f(x)=4/(x-2)^2`

graph{4/(x-2)^2 [-19.46, 20.54, -4.48, 15.52]}