Answer:
[tex]f^{-1}(x)=\pm \sqrt{49x^2-16}[/tex]
Not a function.
Step-by-step explanation:
The given function is;
[tex]f(x)=-\frac{1}{7}\sqrt{16-x^2}[/tex]
Let [tex]y=-\frac{1}{7}\sqrt{16-x^2}[/tex]
Interchange x and y;
[tex]x=-\frac{1}{7}\sqrt{16-y^2}[/tex]
Solve for y;
[tex]-7x=\sqrt{16-y^2}[/tex]
Square both sides
[tex](-7x)^2=(\sqrt{16-y^2})^2[/tex]
[tex]49x^2^2=16-y^2[/tex]
[tex]49x^2^2-16=y^2[/tex]
[tex]y=\pm \sqrt{49x^2-16}[/tex]
The inverse is
[tex]f^{-1}(x)=\pm \sqrt{49x^2-16}[/tex]
[tex]f^{-1}(x)=\pm \sqrt{49x^2-16}[/tex] is not a function because one x-value maps onto to different y-values.
