Question 140

Let Q be the set of all rational numbers and $$f : Q \rightarrow Q$$ be a function defined $$f(x) = \frac{3x - 1}{10}$$ for all $$x \epsilon Q$$. Then $$f^{-1} (5) =$$

Solution

Given $$y\ =\ \frac{\left(3x-1\right)}{10}$$

$$10\ y\ =\ 3x-1$$

Hence$$f^{-1} (x) =$$ = $$\frac{\left(10x+1\right)}{3}$$

Putting x = 5

$$\frac{\left(10x+1\right)}{3}\ =\ \frac{\left(10\cdot5+1\right)}{3}=\frac{51}{3}=17$$

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