Given the function f(x)=3(x+2)-4, how do you solve for the inverse function when x=2?

Aug 19, 2017

${f}^{-} 1 \left(2\right) = 0$

Explanation:

$f \left(x\right) = 3 \left(x + 2\right) - 4$

Let's solve for the inverse of this function

$f \left(x\right) = 3 \left(x + 2\right) - 4$

$y = 3 \left(x + 2\right) - 4$

$x = 3 \left(y + 2\right) - 4$

$x + 4 = 3 \left(y + 2\right)$

$\frac{x + 4}{3} = y + 2$

$\frac{x + 4}{3} - 2 = y$

${f}^{-} 1 \left(x\right) = \frac{x + 4}{3} - 2$

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Now let's solve for $x = 2$

${f}^{-} 1 \left(2\right) = \frac{\left(2\right) + 4}{3} - 2$

${f}^{-} 1 \left(2\right) = \frac{6}{3} - 2$

${f}^{-} 1 \left(2\right) = 2 - 2$

${f}^{-} 1 \left(2\right) = 0$