# Question #bfe8f

May 7, 2017

Given: $f \left(x\right) = {\sin}^{-} 1 \left(7 x - 15\right)$

Substitute ${f}^{-} 1 \left(x\right)$ for every $x$:

$f \left({f}^{-} 1 \left(x\right)\right) = {\sin}^{-} 1 \left(7 {f}^{-} 1 \left(x\right) - 15\right)$

The left side becomes $x$ by definition:

$x = {\sin}^{-} 1 \left(7 {f}^{-} 1 \left(x\right) - 15\right)$

Take the sine of both sides:

$\sin \left(x\right) = 7 {f}^{-} 1 \left(x\right) - 15$

$\sin \left(x\right) + 15 = 7 {f}^{-} 1 \left(x\right)$

Divide both sides by 7:

${f}^{-} 1 \left(x\right) = \frac{\sin \left(x\right) + 15}{7}$

Before one can declare this as the inverse, one must show that $f \left({f}^{-} 1 \left(x\right)\right) = x$ and ${f}^{-} 1 \left(f \left(x\right)\right) = x$:

$f \left({f}^{-} 1 \left(x\right)\right) = {\sin}^{-} 1 \left(7 \left(\frac{\sin \left(x\right) + 15}{7}\right) - 15\right)$

$f \left({f}^{-} 1 \left(x\right)\right) = {\sin}^{-} 1 \left(\sin \left(x\right) + 15 - 15\right)$

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

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

${f}^{-} 1 \left(f \left(x\right)\right) = \frac{\sin \left({\sin}^{-} 1 \left(7 x - 15\right)\right) + 15}{7}$

${f}^{-} 1 \left(f \left(x\right)\right) = \frac{7 x - 15 + 15}{7}$

${f}^{-} 1 \left(f \left(x\right)\right) = \frac{7 x}{7}$

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

Q.E.D.

${f}^{-} 1 \left(x\right) = \frac{\sin \left(x\right) + 15}{7}$

May 7, 2017

${f}^{-} 1 \left(x\right) = \frac{1}{\sin} ^ - 1 \left(7 x - 15\right)$
When $f \left(x\right)$ is equil to the equation and you add it to the power of -1 then you actually just divide 1 by $f \left(x\right)$, thus meaning you should devide 1 with the equation as well. Another answer that would also be correct is just to take the entire equation to the power of -1
${f}^{-} 1 \left(x\right) = {\left({\sin}^{-} 1 \left(7 x - 15\right)\right)}^{-} 1$ wich should give you the same answer.