Question

Which function has a range of y < 3?

Answer 1

Using a graphing tool

Let's graph each of the cases to determine the solution of the problem

case A) $y=3(2^{x})$  

see the attached figure N $1$

The range is the interval--------> (0,∞)

$y> 0$

therefore

the function $y=3(2^{x})$ is not the solution

case B) $y=2(3^{x})$

see the attached figure N $2$  

The range is the interval--------> (0,∞)

$y> 0$

therefore

the function $y=2(3^{x})$ is not the solution

case C) $y=-2^{x}+3$  

see the attached figure N $3$    

The range is the interval--------> (-∞,3)  

$y< 3$

therefore

the function   $y=-2^{x}+3$    is the solution

case D) $y=2^{x}-3$  

see the attached figure N $4$  

The range is the interval--------> (-3,∞)  

$y>-3$

therefore

the function$y=2^{x}-3$ is not the solution

the answer is

$y=-2^{x}+3$

Answer 2

Answer:

The correct option is C.

Step-by-step explanation:

If a function is defined as

$f(x)=a^x$

where, a>0, then the range of the function is greater than 0.

$a^x>0$               .... (1)

Option A: Using inequity (1),

$(2)^x>0$

Multiply both side by 3.

$3(2)^x>0$

$y>0$

The range of first function is y>0. Therefore option A is incorrect.

Option B: Using inequity (1),

$(3)^x>0$

Multiply both side by 2.

$2(3)^x>0$

$y>0$

The range of second function is y>0. Therefore option B is incorrect.

Option C: Using inequity (1),

$(2)^x>0$

Multiply both side by -1.

$-(2)^x$                    $(\text{change the sign of inequality})$

Add 3 on both the sides.

$-(2)^x+3$

$y$

The range of first function is y<3. Therefore option C is correct.

Option D: Using inequity (1),

$(2)^x>0$

Subtract 3 from both the sides.

$2^x-3>-3$

$y>-3$

The range of second function is y>-3. Therefore option D is incorrect.