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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

2 Answers

2 votes

Answer:

B.

Explanation:

answer B is correct for Plato users!!!!

User Zanbaldwin
by
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4 votes

Answer:

C.Restricted domain :
x\geq -4,
f^(-1)(x)=-4+√(x-3)

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=√(y-3)


x=√(y-3)-4

Hence,
f^(-1)(x)=-4+√(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 .

User Suave Nti
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5.1k points