find (g/f)(x) if f(x)=v x^2-1 g(x)=v x-1

Question

the v is a radical sign 

Answer 1

Nathalie,

I don't recall ever seeing (g/f)(x) before. Were you trying to show g(f(x))?

 

If your typing was correct, then (g/f)(x) would be the same as g(x)/f(x) or:

 

√(x-1)/√(x²-1)           Now recall that √16/√4 = √(16/4), so we can rewrite the expression as:

√[(x-1)/(x²-1)]           So even though we don't have a square root for either binomial, we can reduce

 

√{(x-1)/[(x-1)(x+1)]} = √[1/(x+1)] = 1/√(x+1)

If we need to rationalize, then we get √(x+1)/(x+1)

 

 

Now the other problem: g(f(x)) just means plugging in the f function for x in the g function.

 

g(f(x)) = √{[√(x²-1)]-1}  This one cannot be simplified.

 

I hope one of these was helpful.

Answer 2

Hi Nathalie,

 

To find (g/f)(x)

f(x) =√(x² - 1)

g(x)=√x^-1

 

(g/f)(x) = g(x)/f(x)

            =√x^-1/(√(x² - 1))

            =1/√x(x² -1)

I am not sure if you intended

g(x) = √(x-1) if so, then

(g/f)(x) = g(x)/f(x)

            =√(x -1)/(√(x² -1))

            =√(x-1)/[(√(x - 1)(x + 1)]

            =√(x-1)/[√(x -1)√(x+1)]  the √(x-1) at the top cancels the one at the bottom

            = 1/√(x+1)