Suppose `f(x)=(x+1)^(2) " for " x ge -1. ` If g(x) is the function whose graph is the reflection of the graph of f(x) with respect to the line y = x, then g(x) equals

Options

A. `-sqrtx-1, x ge0`
B. `(1)/((x+1)^(2)), x gt -1`
C. `sqrt(x+1), x ge -1`
D. `sqrtx-1, x ge0`

Answer

D

Explanation

For all `x ge -1` we have
`f(x)=(x+1)^(2) ge 0`
Therefore, `f:[-1,oo]to [0,oo)"given by "f(x)=(x+1)^(2)` is a bijection and hence invertible.
It is given that g(x) is the function whose graph is the reflection of the graph, of f(x) with respect to the line y=x. This means that g(x) is the inverse of f(x).
`fog(x)=x "for all "x ge0`
`Rightarrow f(g(x))=x "for all "x ge0`
`Rightarrow {(g(x)+1}^(2)=x " for all "x ge0`
`Rightarrow (g(x)+1=pmsqrtx " for all "x ge0`
`Rightarrow (g(x)=-1pm+sqrtx " for all "x ge0`
`Rightarrow g(x)=-1+sqrtx " for all "x ge 0" "[therefore g(x) ge-1]`