How to Calculate the function of an interval? I need to calculate $f((1,4])$ for the function 
$$f(x)=x^2-4x+3.$$
The answers I can choose from are:
a) [0,3]
b) [-1,0)
c) (0,3]
d) [-1,3]
e) (-1,0)
f) (0,3)
Can someone guide me? It may be something simple but I don't know how to proceed. Thank you very much!
 A: We want to get a good grasp of $f(x)$. One way I would recommend is to draw the graph $y=f(x)$. (If necessary, you might have some software do the drawing, but don't necessarily trust the result.)  Regrettably, I will have to do things without a picture.
By completing the square, we see that $f(x)=(x-2)^2-4+3=(x-2)^2-1$. So the curve $y=f(x)$ is a parabola. Now we can trace out $f(x)$ as $x$ travels from $1$ to $4$.
At $x=1$ (which is not in the interval $(1,4]$), we have $f(x)=0$. Then as $x$ travels from $1$ to $2$, $f(x)$ decreases, until it reaches $-1$ at $x=2$. So the vertex of the parabola is at $(2,-1)$. Then, as $x$ increases from $2$ to $4$, $(x-2)^2-1$ increases from $-1$ to $3$. 
So all values from $-1$ to $3$, inclusive, are taken on by $f(x)$, as $x$ travels over the interval $(1,4]$. The answer is therefore $[-1,3]$.
A: Note that $f(x)$ can be factorized as $(x-3)(x-1)$, so that the zeros are in 3 and 1. $f$ is negative between 1 and 3. The minimum is at 2, at which the value is -1. Now you should be able to draw the parabola which is the graph of $f$ and find the maximum of $f$ in $[1,4]$.
A: The graph of your function $f$ is a parabola that opens up. Its vertex has $x$-coordinate $x={-(-4)\over 2\cdot 1}=2$ (the vertex of the graph of $y=ax^2+bx+c$ has $x$-coordinate $-b\over 2a$).  So, evaluate $f(2)$ (this gives the minimum value over $(1,4]$), $f(1)$ and $f(4)$.  From those values you can determine $f((1,4])$. 
You can save even more time by exploiting symmetry: since the line through the vertex of a parabola is a line of symmetry, the maximum value of $f$ over $(1,4]$ is $f(4)$ ($2$ is closer to $1$ than to $4$).
