1. ## Range of Functions

Find the range of f(x) = 2x^2 - x - 1.

How is this done algebraically?

2. ## Re: Range of Functions

We observe that the given function is a quadratic, whose graph opens upwards since the coefficient on the squared term is positive. So, we know there is no upper bound on the range. Thus the lower bound will occur on the axis of symmetry:

$\displaystyle x=-\frac{-1}{2(2)}=\frac{1}{4}$

$\displaystyle f_{\min}=f\left(\frac{1}{4}\right)=?$

3. ## Re: Range of Functions

Originally Posted by MarkFL
We observe that the given function is a quadratic, whose graph opens upwards since the coefficient on the squared term is positive. So, we know there is no upper bound on the range. Thus the lower bound will occur on the axis of symmetry:

$\displaystyle x=-\frac{-1}{2(2)}=\frac{1}{4}$

$\displaystyle f_{\min}=f\left(\frac{1}{4}\right)=?$
What do you mean by upper and lower bound?

4. ## Re: Range of Functions

Originally Posted by harpazo
What do you mean by upper and lower bound?
The global maximum/minimum.

5. ## Re: Range of Functions

Originally Posted by MarkFL
The global maximum/minimum.
Can you show me the meaning of upper and lower bound using a graph?

6. ## Re: Range of Functions

Originally Posted by harpazo
Can you show me the meaning of upper and lower bound using a graph?
For a parabola opening upwards, the minimum is at the vertex, and there is no maximum as the function has no upper bound, that is, it can be made as large as we want the farther we take an x value from the axis of symmetry.

7. ## Re: Range of Functions

Originally Posted by MarkFL
For a parabola opening upwards, the minimum is at the vertex, and there is no maximum as the function has no upper bound, that is, it can be made as large as we want the farther we take an x value from the axis of symmetry.
Ok. If the coefficient of the x^2 term is positive, the parabola opens upward. Thus, we have a minimum at the vertex.

If the coefficient of the x^2 term is negative, the parabola opens downward. Thus, we have a maximum at the vertex.

Is this right?

8. ## Re: Range of Functions

Originally Posted by harpazo
Ok. If the coefficient of the x^2 term is positive, the parabola opens upward. Thus, we have a minimum at the vertex.

If the coefficient of the x^2 term is negative, the parabola opens downward. Thus, we have a maximum at the vertex.

Is this right?
Yes, that's correct.

9. ## Re: Range of Functions

Originally Posted by MarkFL
Yes, that's correct.
Ok. I have a somewhat understanding of max/min.