# Math Help - Intervals of Increase and Decrease

1. ## Intervals of Increase and Decrease

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

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

the book has the answer:

f(x) is increasing for x>2;f(x) is decreasing for -1<x<1

I would just like to know how the got the answer step by step because the book does not give a great example. Thanks!

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

f ' (x) = 2x - 4 which is simply a line yint - 4 and x int 2 with slope 2
The only 0 is 2x-4=0 or x=2

It is fairly easy to see then f ' > 0 if x >2 therefore f is incr
f' < 0 if x < 2 therefore f is decr

Don't know where the -1 < x< 1 comes from but it isn't correct

3. Originally Posted by lisa1984wilson
Find the intervals of increase and decrease for the given function:

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

the book has the answer:

f(x) is increasing for x>2;f(x) is decreasing for -1<x<1

I would just like to know how the got the answer step by step because the book does not give a great example. Thanks!
You know that a function is incresing when the slope of a line tangent to the curve is positive, and decreasing when slope is negative. Well, the slope of a curve at any given point is nothing more than the first derivative of the function in question evaluated at that point. All we've gotta do is find the first derivative and see when it's positive or negative. Like so:

By the power rule $\frac{d}{dx}(x^2-4x+5)=2x-4$

Now, to find when its greator than zero

$2x-4>0\Longleftrightarrow{x>2}$

Similarily for less than zero

$2x-4<0\Longleftrightarrow{x<2}$

And the slope = 0, of course, when x=2.

So, therefore, since the first derivative of the function is nothing more than a formula for finding slope, we know that the function is increasing when x>2, decreasing when x<2, and at a minimum when x=2.

Oh, and by the way, either three things happened here:

1.You miscopied the answer out of the book.
2.The book is wrong.
3.You didn't provide all of the necessary information to the forum.

(what im trying to say, is that you were right to question the answer givin.)

The reason I say this is because it is obvious that the graph of this particular parabola opens upward infinitely in both directions, and has one and only one minimum.