# Thread: Need help with mins and maxes.

1. ## Need help with mins and maxes.

What do I do if I have more than one maxes or mins?

For example in this problem.

Let $f(x)=-x^2+3x$ on the interval $[1,3]$. Find the Absolute Maximum and absolute minimum of f(x) on this interval.

I know that the min occurs at x=3.

But I don't understand what to put if I have more than one (in this case) max.

2. Once you have a list of all the posible maximums and minimums (i.e., critical values), plug each one into the original function. Whichever one yields the greatest value is the absolute maximum; whichever one yields the smallest value is the absolute minimum.

3. Originally Posted by drumist
Once you have a list of all the posible maximums and minimums (i.e., critical values), plug each one into the original function. Whichever one is the greatest is the absolute maximum; the least is the absolute minimum.

Are you saying to use it's derivative first?

4. Originally Posted by Zanderist
What do I do if I have more than one maxes or mins?

For example in this problem.

Let $f(x)=-x^2+3x$ on the interval $[1,3]$. Find the Absolute Maximum and absolute minimum of f(x) on this interval.

I know that the min occurs at x=3.

But I don't understand what to put if I have more than one (in this case) max.
Start by drawing the graph over the given interval. Label the coordinates of the turning point and the end points. Now, do you know and understand the definition of an absolute maximimum and an absolute minimum?

5. Originally Posted by mr fantastic
Start by drawing the graph over the given interval. Label the coordinates of the turning point and the end points. Now, do you know and understand the definition of an absolute maximimum and an absolute minimum?
yes, it's the highest point and the lowest point in the graph. Since the graphs go on for infinite in both directions (in this case at least because a square root and log would be much different) we use the intervals listed to us and we work within those parameters.

The critical point is when it intercepts the x-axis. To find that, you take the derivative and solve for 0. I was able to solve this problem by doing that, though I still have trouble with other graphs.

6. Originally Posted by Zanderist
Are you saying to use it's derivative first?
Have you covered how to use the derivative to find minimum and maximum values of a function? That is basically what you need to do here. I presumed this was the intended method since you are posting in the calculus forum. But I have no idea what you have covered in your coursework or what method your teacher intends for you to use.

However, there are other ways to solve it as well. For one thing, the function is a parabola. Since the coefficient of the $x^2$ term is negative, the parabola opens downward, which means its maximum point is at the vertex. Also since the graph will always be decreasing when you move away from the vertex, its minimum point will be one of the endpoints of the interval. You could evaluate the function at each endpoint to figure out which one is the minimum. (Sounds like you already know the minimum though.)

7. Originally Posted by drumist
Have you covered how to use the derivative to find minimum and maximum values of a function? That is basically what you need to do here. I presumed this was the intended method since you are posting in the calculus forum. But I have no idea what you have covered in your coursework or what method your teacher intends for you to use.
I don't think I know this method.

8. Originally Posted by Zanderist
I don't think I know this method.
In that case, you are probably supposed to use the algebra approach.

The vertex of a parabola whose equation is of the form $y=ax^2+bx+c$ is located at $x=-\frac{b}{2a}$ .