Answer:
We have that [tex]f(a) = f(b)[/tex] if [tex]a = \pm b[/tex], which does not respect the condition for the existence of an inverse function, which means that [tex]f^{-1}(x)[/tex] does not exist for this function.
Step-by-step explanation:
Existence of an inverse function:
An inverse function will exist if: f(a) = f(b) only if a = b.
In this question:
[tex]f(a) = -\frac{1}{7}\sqrt{16 - a^2}[/tex]
[tex]f(b) = -\frac{1}{7}\sqrt{16 - b^2}[/tex]
[tex]-\frac{1}{7}\sqrt{16 - a^2} =-\frac{1}{7}\sqrt{16 - b^2}[/tex]
[tex]\sqrt{16 - a^2} = \sqrt{16 - b^2}[/tex]
[tex](\sqrt{16 - a^2})^2 = (\sqrt{16 - b^2})^2[/tex]
[tex]16 - a^2 = 16 - b^2[/tex]
[tex]a^2 = b^2[/tex]
[tex]a = \pm b[/tex]
We have that [tex]f(a) = f(b)[/tex] if [tex]a = \pm b[/tex], which does not respect the condition for the existence of an inverse function, which means that [tex]f^{-1}(x)[/tex] does not exist for this function.
