Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
It helps to make a drawing – Egbert May 23 '12 at 18:44
As Egbert said you can draw a picture and look at the $y$ axis and see which points on it have a corresponding $x$ value in that interval:^2%E2%88%924x%2B3%2C+x%3D1+to+4 – Keivan May 23 '12 at 18:51
up vote 2 down vote accepted

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 $[-1,3]$.

share|cite|improve this answer
Thank you so much! Been having trouble with this type of exercise :)! – Grozav Alex Ioan May 23 '12 at 19:04

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 $[1,4]$.

share|cite|improve this answer

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 $4$).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.