Critical numbers: intervals of incr/decr of f(x) = x^2-4x+5

[charlenej]

The question is:

Find the intervals of increase and decrease for the given function:

. . .f(x) = x^2 - 4x + 5

I know the first step is to set find the derivative of the equation, and set it to zero, which is:

. . .0 = 2x - 4

Then I solved for x:

. . .x = 2

And now, this is where I'm lost. I don't know how to find the critical numbers to solve the rest of the problem.

 

[galactus]

A critical point is any value of x in the domain of f at which f'(x)=0 or where it's not differentiable.

Your differentiable everywhere.

You found a critical point with x=2.

Relative extrema occur at critical points.

Check to see if it's a max or min. by the 2nd derivative test.

\(\displaystyle f''(x)=2\)

If f''(x) > 0, then relative minimum. Which you can see by graphing.

Your critical point is at (2,1)

If f''(x) > 0 on an interval [a,b], then f is concave up.

A relative minimum occurs at those critical points where the sign of f'(x) changes from negative to positive moving in the positive x-direction.

If f'(x)>0 for every value of x in (a,b), then f is increasing.
If f'(x)<0 for every value of x in (a,b), then f is decreasing.
If f'(x)=0 for every value of x in (a,b), then f is constant.

Since f'(x) < 0 if \(\displaystyle {-\infty}<x<2\)

and f'(x)>0 if \(\displaystyle 2<x<{\infty}\)

f is continuous at x=2, therefore,

f is decreasing on \(\displaystyle ({-\infty},2]\)

f is increasing on \(\displaystyle [2,{\infty})\)