Functions:
[tex]h(x)=x^2+1[/tex][tex]h^{-1}(x)=\pm\sqrt[]{x-1}[/tex]
The inverse function's graph has to be symmetrical across the y = x line, if we graph this line with the previous graphs:
where the red line is h(x), the blue and green are the possibilities:
[tex]h^{-1}(x)=+\sqrt[]{x-1}[/tex]
...and...
[tex]h^{-1}(x)=-\sqrt[]{x-1}[/tex]
...respectively. And the purple line is the line y = x.
As the inverse functions are symmetrical across the y = x line, then we know our functions are inverse.
Answer:
[tex]h^{-1}(x)=\pm\sqrt[]{x-1}[/tex]
• Graph
Reason why it is inverse: the functions are symmetrical across the y = x line.
