4.2: Extreme Values

Finding the derivative is one of my favorite parts,
they're used to find extrema on various charts.
The critical points are easy to find,
they're the horizontal tangent line.


Problem:

1. Find the critical point(s) at f(x)=9x^2-5x+4.
f'(x)=18x-5
f'(x)=18x-5=0
f'(x)=18x=5
x=5/18.

f(x)=9(5/18)^2-5(5/18)+4 = 3.305
y=3.305

The critical point for this function is ((5/18),3.305)