Question

1. Which of the following functions has a range of f(x) <3? a. f(x) = 2x + 3b. f(x) = 3x + 8c. f(x) = -x2 + 3d. f(x) = (x - 3)2 + 3ABaСD

Answer 1

we can check each option using the limit number f(x)=3 and a number out of the range(4)

must have a value for the limit number but a mistake when we use 4

A.

f(x)=3

`\begin{gathered} 3=2^x+3 \\ 3-3=2^x \\ 0=2^x \end{gathered}`

any value of x will result in 0

then A is wront because dont have a solution for the limit number

B.

f(x)=3

`\begin{gathered} 3=3x+8 \\ 3-8=3x \\ -5=3x \\ x=-(5)/(3) \end{gathered}`

has a solution for f(x)=3

f(x)=4

`\begin{gathered} 4=3x+8 \\ 4-8=3x \\ -4=3x \\ x=-(4)/(3) \end{gathered}`

has a solution for a value out of the range then option B is wrong

C.

f(x)=3

`\begin{gathered} 3=-x^2+3 \\ 3-3=-x^2 \\ 0=-x^2 \\ -x=\sqrt[]{0} \\ -x=0 \\ x=0 \end{gathered}`

has a solution for f(x)=3

f(x)=4

`\begin{gathered} 4=-x^2+3 \\ 4-3=-x^2 \\ 1=-x^2 \\ x^2=-1 \\ x=\sqrt[]{-1} \end{gathered}`

root of negative number doesnt have real solution , then this fuction dont have solution for a number out of the range

Option C is RIGHT because has solution on the limit number f(x)=3 but not a solution for a number out of range