Respuesta :
[tex] \\(a)\\ \text{Consider the function, }\\ \\ f(x)=\sqrt{x-10}\\ \\ \text{let }y=f(x)\\ \\ \text{so in order to find the inverse function, first we interchange x and y.}\\ \text{so we have}\\ \\ x=\sqrt{y-10}\\ \\ \text{now we solve for y and that will give the required inverse function.}\\ \\ \text{square both sides to get} \\ \\ x^2=(\sqrt{y-10})^2\\ \\ \Rightarrow x^2=y-10\\ \\ \Rightarrow y=x^2+10\\ \\ \text{so the inverse function is }\\ \\ f^{-1}(x)=x^2+10. [/tex]
[tex] \\(b)\\ \text{The graphs of f and }f^{-1} \text{ are shown below:}\\ \\ (c)\\ \text{from the graph of f(x), we can see that}\\ \\ \text{Domain}=[10,\infty)\\ \\ \text{Range}=[0,\infty)\\ \\ \text{and from the graph of inverse function }f^{-1}(x), \text{ we can see that}\\ \\ \text{Domain}=[0, \infty)\\ \\ \text{Range}=[10, \infty) [/tex]

Answer:
What are the roots of f(x) = x2 – 48?
–48 and 48
–24 and 24
Negative 8 StartRoot 3 EndRoot and 8 StartRoot 3 EndRoot
Negative 4 StartRoot 3 EndRoot and 4 StartRoot 3 EndRoot
THE ANSWER IS D