Skip to main contentSkip to solution

Let f(x)=x(2x−1)f(x) = \dfrac{x}{(2x-1)} and g(x)=x(x−1)g(x) = \dfrac{x}{(x-1)}. Then, the domain of the function h(x)=f(g(x))+g(f(x))h(x) = f(g(x)) + g(f(x)) is all real numbers except

Solution

✅ Correct Option: 2

g(x)g(x) is undefined at x=1x = 1.

f(x)f(x) is undefined at x=12x = \dfrac{1}{2}.


For f(g(x))f(g(x)), the denominator 2g(x)−1=02g(x) - 1 = 0 gives:

2⋅xx−1−1=02 \cdot \dfrac{x}{x-1} - 1 = 0

2x−(x−1)x−1=0\dfrac{2x - (x-1)}{x-1} = 0

x+1x−1=0\dfrac{x+1}{x-1} = 0

x=−1x = -1


For g(f(x))g(f(x)), the denominator f(x)−1=0f(x) - 1 = 0 gives:

x2x−1−1=0\dfrac{x}{2x-1} - 1 = 0

x−(2x−1)2x−1=0\dfrac{x - (2x-1)}{2x-1} = 0

−x+12x−1=0\dfrac{-x+1}{2x-1} = 0

x=1x = 1 (already excluded)


The excluded values are −1,12-1, \dfrac{1}{2}, and 11.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question