SOLUTION: Find necessary and sufficient conditions on a, b, c and d to ensure that f(x) = (ax + b)/(cx + d) is a constant function. [Hint: what is the derivative of a constant function?]

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Find necessary and sufficient conditions on a, b, c and d to ensure that f(x) = (ax + b)/(cx + d) is a constant function. [Hint: what is the derivative of a constant function?]      Log On


   



Question 1134916: Find necessary and sufficient conditions on a, b, c and d to ensure that
f(x) = (ax + b)/(cx + d)
is a constant function. [Hint: what is the derivative of a constant function?]

Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Algebraically, if the rational function is a constant function, k, then...

%28ax%2Bb%29%2F%28cx%2Bd%29+=+k
ax%2Bb+=+k%28cx%2Bd%29
ax%2Bb+=+kcx%2Bkd
a+=+kc and b+=+kd

This result can be expressed in many ways, including

a%2Fb+=+c%2Fd

Using calculus to find the derivative and find the conditions that make the derivative zero....

f%28x%29+=+%28ax%2Bb%29%2F%28cx%2Bd%29


The condition for the derivative to be zero is

ad-bc=0

which is equivalent to the earlier forms.

Answer by ikleyn(52784) About Me  (Show Source):
You can put this solution on YOUR website!
.
From the first glance to the problem, the answer is obvious:


    The necessary and sufficient condition is proportionality of the pairs (a,b) and (c,d).


To get a formal proof, you can take the derivative of the function.


The numerator of the derivative function then is 


    a*(cx + d) - (ax + b)*c = (ac - ac)*x + ad - bc = ad - bc.


The function (the original rational function / (fraction) ) is a constant if and only if the numerator of the derivative is identically zero, 

which leads to the equality  ad - bc = 0.


In turn, it means that  ad = bc,  or, equivalently,  a%2Fb = c%2Fd.


It is precisely the same condition as proportionality of the pairs (a,b)  and  (c,d).

Solved and answered.