When $f(x) = 3x-4$ and $g(x) = \frac{5}{3-x}$,
Question 1: Find the value of x for which fg(x) = 5
Question 2: Show that the equation $f^{-1}(x) = g^{-1}(x)$ has no real roots.
I understand that in question 1 it simply is asking to plug 5 into the x and then solve for fg(x) =
5/(3-5) = -2.5
3(-2.5) – 4 = -11.5 (this is all right so far?)
I don't understand question 2, or how exactly I should answer it. How do I work out that there are no real roots? This is how the question is asked. Obviously the functions have to first be in the inverse forms.
I'm not an Einstein when it comes to working out roots – so if possible – please explain each step taken so that I may understand this type of question in the future.
Many thanks,
Student909.
Best Answer
$$f(x)=3x-4$$ $$x=\frac{f(x)+4}3$$ $$f^{-1}(x)=\frac{x+4}3$$ $$g(x)=\frac{5}{3-x}$$ $$x=3-\frac{5}{g(x)}$$ $$g^{-1}(x)=3-\frac{5}{x}$$ $$f^{-1}(x)-g^{-1}(x)=\frac x3+\frac43-3+\frac 5x=\frac{x^2-5x+15}{3x}$$ The discriminant of $x^2-5x+15$ is negative. So, $x^2-5x+15\ne0$ for all $x\in\mathbb R$.
Thus, $$f^{-1}(x)-g^{-1}(x)\ne0$$ for all $x\in\mathbb R$.
As a result, $$f^{-1}(x)=g^{-1}(x)$$ has no real roots.