Use the image to answer the question.

Question

A coordinate plane's axes range from negative 4 to 4, both by 1-unit increments. A dotted curve labeled g of x and a solid curve labeled f of x are plotted.

Compare the given graphs of f(x)=2x−−√
and g(x)=2x−−√3
and determine which of the following is true.

(1 point)
Responses

The graphs have different domains.
The graphs have different domains.

Both graphs go through (−1,−1)
, (0,0)
, and (1,1)
.
Both graphs go through left parenthesis negative 1 comma negative 1 right parenthesis , left parenthesis 0 comma 0 right parenthesis , and left parenthesis 1 comma 1 right parenthesis .

They are both decreasing on their domains.
They are both decreasing on their domains.

When x>1
, the function g(x)=2x−−√3
is greater than f(x)=2x−−√
.

Answer 1

To compare the functions $f(x)=\sqrt{2x}$ and $g(x)=\sqrt{2x}+3$, let's analyze each statement one by one.

1. The graphs have different domains.

   • Both functions $f(x)=\sqrt{2x}$ and $g(x)=\sqrt{2x}+3$ only exist for $x\ge 0$ since the square root function is only defined for non-negative numbers. Thus, their domains are the same: $[0,\mathrm{\infty })$. This statement is false.

2. Both graphs go through (−1,−1), (0,0), and (1,1).

   • For $f(x)=\sqrt{2x}$:
     • At $x=0$: $f(0)=\sqrt{2(0)}=0$ (so, it goes through (0, 0)).
     • At $x=1$: $f(1)=\sqrt{2(1)}=\sqrt{2}$, approximately 1.414 (so, it does not go through (1, 1)).
     • At $x=-1$: $f(-1)$ is not defined (so, it does not go through (-1, -1)).
   • For $g(x)=\sqrt{2x}+3$:
     • At $x=0$: $g(0)=\sqrt{2(0)}+3=3$.
     • At $x=1$: $g(1)=\sqrt{2(1)}+3=\sqrt{2}+3$, approximately 4.414 (not (1, 1)).
     • At $x=-1$: $g(-1)$ is also not defined.
   • Thus, this statement is false.

3. They are both decreasing on their domains.

   • For $f(x)=\sqrt{2x}$ and $g(x)=\sqrt{2x}+3$, both functions are increasing as $x$ increases. Therefore, this statement is false.

4. When $x>1$, the function $g(x)=\sqrt{2x}+3$ is greater than $f(x)=\sqrt{2x}$.

   • For $x>1$:
     • Since $g(x)=\sqrt{2x}+3$ adds a constant (3) to $f(x)$, it is clear that $g(x)$ is always greater than $f(x)$ for $x>1$. This statement is true.

Based on the evaluations, the correct answer is:

When x>1, the function g(x)=√(2x)+3 is greater than f(x)=√(2x).