Answer:-
The two inverses are,
[tex]y = \sqrt {\frac{x+4}{9}}[/tex]
and, [tex]y = - \sqrt {\frac{x+4}{9}}[/tex]
Step-by-step calculation:-
[tex]y = 9 \times x^{2} - 4 [/tex]
⇒[tex]x^{2} = \frac{y+4}{9}[/tex]
⇒[tex]x = \mp \sqrt {\frac{y+4}{9}}[/tex]
So, the two inverses are,
[tex]y = \sqrt {\frac{x+4}{9}}[/tex]
and, [tex]y = - \sqrt {\frac{x+4}{9}}[/tex]
Answer:
[tex]f^{-1}(x)=\pm \frac{\sqrt{x+4} }{3}[/tex]
Step-by-step explanation:
In order to find the inverse of [tex]y=9x^2-4[/tex] we need to follow the next steps:
Step 1: Solve for x
Add 4 to both sides:
[tex]y+4=9x^2-4+4\\y+4=9x^2[/tex]
Divide by 9 from both sides:
[tex]\frac{y+4}{9}=\frac{9x^2}{9} \\\frac{y+4}{9}=x^2[/tex]
Square root from both sides:
[tex]\sqrt{x^2}=\pm \sqrt{\frac{y+4}{9}} \\x=\pm \frac{\sqrt{y+4} }{3}[/tex]
Step 2: Replace every x with a y and replace every y with an x.
[tex]y=\pm \frac{\sqrt{x+4} }{3}[/tex]
So:
[tex]f^{-1}(x)=\pm \frac{\sqrt{x+4} }{3}[/tex]
Step 3: Verify your work by checking that:
[tex](f \hspace{3}o\hspace{3}f^{-1})(x)=x\\(f^{-1}\hspace{3}o\hspace{3}f)(x)=x[/tex]
[tex](f \hspace{3}o\hspace{3}f^{-1})(x)=x=9(\frac{\sqrt{x+4} }{3})^2 -4=x+4-4=x[/tex]
[tex](f^{-1}\hspace{3}o\hspace{3}f)(x)=x=\frac{\sqrt{9x^2-4+4} }{3} =\frac{\sqrt{9x^2} }{3} =\frac{3x}{3} =x[/tex]
