MHB Graph y=g(x): Sketch for [0,5] \implies R

[Katsa333]

I don't really know how to start with this question. Please help?

For the function g: [0,5] \implies R, g(x)=(x+3)/(2) (R=Real Numbers)
sketch the graph of y=g(x)
I don't know how the [0,5] \implies R changes the graph.

 

[I like Serena]

I don't really know how to start with this question. Please help?

For the function g: [0,5] \implies R, g(x)=(x+3)/(2) (R=Real Numbers)
sketch the graph of y=g(x)
I don't know how the [0,5] \implies R changes the graph.

Hi Katsa333! Welcome to MHB! (Wave)

Properly we have the function $g: [0,5] \to \mathbb R$ given by $g(x)=\frac{x+3}{2}=\frac 12 x + \frac 32$.
The first part does not change the graph, other than defining its domain [0,5], meaning it begins at x=0 and ends at x=5.
The second part is the equation of a line that slopes up by $\frac 12$ when we move $1$ to the right.
And it intercepts the y-axis at $y=\frac 32$.

Now what will the graph look like? (Wondering)

 

[HOI]

Ah! I wondered what "[0, 1] implies R" meant! I like Serena is, correctly I think, taking it to mean that f is a function from [0, 1] to R.

Katsa333, "→" here is NOT "implies", it is simply "to" or "goes to". As I like Serena said, the graph of the equation y= (x+ 3)/2 is a straight line, with slope 1/3 and y-intercept 3/2. Restricting x to [0, 1] means that the graph is only the part of that line that lies above [0, 1] on the x-axis. It is the line segment with endpoints (0, 3/2) and (1, 2).