Ok. Me and my class is going to do a 30-minute algebra test and my teacher gave us some practice sheets. One of them has this problem:Given f(x) = x^2-6x+4. Find the minimum and maximum of f(x), given that 1 <= x <= 4. HELP ME!
Yeah, that really helps. Thanks
f(x) = x^2 - 6x + 4
= (x^2 -6x +9) -9 +4
So the vertex is at the point V(3,-5) and the parabola opens up. So y = -5 is a "local" minimum.
Now all we have to do is check the endpoints of the interval. f(1) = -1 and f(4) = -4
Here is a graph in the interval 1<=x<=4
So we see that the max value is f(1) = -1 and the min value is at the vertex = -5
The key in answering questions about max/min values of a function inside a closed interval is that we need to check the y-values at the endpoints of the interval as well as at the vertex. We do this because the y-values at the endpoints may or may not be larger/smaller than the y-value of the vertex (depending on the equation and the interval). The graph is not needed to answer the question but it does allow you to fully understand why we are doing the algebra we are doing. Hope this helps.
Yes,it is. Thanks. Just a bonus question: How did you figure it out without calculus. graphs and only algebra?
Check the y-values at x=1, the vertex, x=4 and you will have your answer answer.
Oh, i forgot. We can't use graphs too. Sry. And I do know how to complete the square.
Do you know how to complete the square to find the vertex? And then use that vertex to help you sketch the parabola?