Question

-Exponential and Logarithmic Functions- If h(x) = x² + 1...

Answer 1

Functions:

`h(x)=x^2+1`
`h^(-1)(x)=\pm\sqrt[]{x-1}`

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:

`h^(-1)(x)=+\sqrt[]{x-1}`

...and...

`h^(-1)(x)=-\sqrt[]{x-1}`

...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:

`h^(-1)(x)=\pm\sqrt[]{x-1}`

• Graph

Reason why it is inverse: the functions are symmetrical across the y = x line.