Question

The function f(x)=(x+4)^2+3 is not one-to-one. Identify a restricted domain that makes the function one-to-one, and find the inverse function.
A. restricted domain: x<= -4; f^-1(x)=-4+ sqrt x-3

B. restricted domain: x>=-4; f^-1 (x)=-4 - sqrt x+3

C. restricted domain: x>= -4; f^-1(x) =-4+ sqrt x-3

D. restricted domain: x<= -4; f^-1(x)= -4+ sqrt x+3

Answer 1

Answer:

C.Restricted domain :$x\geq -4$, $f^{-1}(x)=-4+\sqrt{x-3}$

Step-by-step explanation:

We are given that a function is not one - to-one.

$f(x)=(x+4)^2+3$

Suppose $y=(x+4)^2+3$

$y-3=(x+4)^2$

$x+4=\sqrt{y-3}$

$x=\sqrt{y-3}-4$

Hence, $f^{-1}(x)=-4+\sqrt{x-3}$

We know that domain of f(x) is converted into range of $f^{-1}(x)$ and range of f(x) is converted into domain of $f^{-1}(x)$.

Substitute x=3 then we get

$f^{-1}(x)=-4$

Domain of $f^{-1}(x)=[3,\infty)$

Range of $f^{-1}(x)=[-4,\infty)$

Domain of f(x)=$[-4,\infty)$

Restricted domain :$x\geq -4$

Hence, restricted domain of f(x) that makes the function one-to-one .

Answer 2

Answer:

B.

Step-by-step explanation:

answer B is correct for Plato users!!!!