Question

Find critical points and extreme values on the interval [-1,5]:

f(x)=|x^2-4x+3|

Answer 1

You know what parabolas look like.

(x-1)(x-3) is negative between the roots, so this graph has a bump in the middle.

Since the axis of symmetry is x=2, f(-1) = f(5) = 8
f(2) = 1

so, the max value is 8 and the minima are 0.

f(2) is also a local max, since it is the vertex of the parabola. (also, f'(2) = 0!)

Now that you have it all worked out, take a peek at the graph:

http://www.wolframalpha.com/input/?i=%7Cx%5E2-4x%2B3%7C++for+-1+%3C%3D+x+%3C%3D+5