# How do you find the intervals of increasing and decreasing using the first derivative given #y=5-abs(x-5)#?

##### 1 Answer

# x in (-oo, 5) => # increasing

# x in (5,oo) \ \ \ \ \=> # decreasing

#### Explanation:

We have:

# y=5-|x-5| #

It always helps to draw a graph, we cam probably answer the question fully from the graph, at the very least it can help guide the solution.

graph{5-|x-5| [-8.42, 11.58, -4, 6]}

From the graph we can see

Let us see if was can establish this same solution analytically. The definition of an increasing or decreasing function is that

# { (y' < 0),(y' > 0) :} => {: (y " is decreasing"), (y " is increasing") :} #

It is the modulus in the function that will complicate analysis, so let's remove that using its' definition, and break the function into analytical segments. By definition;

# |x| = { (-x \ \ \ , x<0), (0, x=0), (x, x>0) :} #

And so:

# |x-5| = { (-(x-5) \ \ \, x-5<0), (0, x-5=0), (x-5, x-5>0) :} #

# \ \ \ \ \ \ \ \ \ \ = { (-(x-5) \ \ \, x<5), (0, x=5), (x-5, x>5) :} #

And finally:

# :. \ \ -|x-5| = { ((x-5) \ \ \, x<5), (0, x=5), (-(x-5), x>5) :} #

# :. 5-|x-5| = { (5+(x-5) \ \ \, x<5), (5+0, x=5), (5-(x-5), x>5) :} #

# :. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y = { (x \ \ \, x<5), (5, x=5), (10-x, x>5) :} #

And we can now differentiate to get:

# y' = { (1 \ \ \, x<5), ("undefined", x=5), (-1, x>5) :} #

And so the derivative condition are:

#y'>0 => x<5 => x in (-oo, 5) => # increasing

#y'<0 => x>5 => x in (5,oo) \ \ \ \ \=> # decreasing

which gives intervals consistent with our graphical analysis.