Question 8

# Let $$f(x) = \frac{x^2 + 1}{x^2 - 1}$$ if $$x \neq 1, -1,$$ and 1 if x = 1, -1. Let $$g(x) = \frac{x + 1}{x - 1}$$ if $$x \neq 1,$$ and 3 if x = 1.What is the minimum possible values of $$\frac{f(x)}{g(x)}$$ ?

Solution

$$\frac{f\left(x\right)}{g\left(x\right)}=\frac{\left(x^2+1\right)}{x^2-1}\cdot\frac{\left(x-1\right)}{x+1}=\frac{\left(x^2+1\right)}{\left(x+1\right)^2}$$

This function is definitely greater than 0

$$let\ y=\frac{\left(x^2+1\right)}{\left(x+1\right)^2}$$

=> $$x^2\left(y-1\right)+2yx+\left(y-1\right)=0$$ which is quadratic in x

Disctiminant should be greater than 0

$$4y^2-4\left(y-1\right)^2\ge0$$

=> y>=1/2

When x =1, f(x)/g(x) = 1/3

Hence either the value should be greater than 1/2 or should be equal to 1/3