The function #f:f(x)=-|x|+1# is decreasing in the interval...?

2 Answers
May 10, 2018

Decreasing on #(0, oo)#

Explanation:

To determine when a function is increasing or decreasing, we take the first derivative and determine where it is positive or negative.

A positive first derivative implies an increasing function and a negative first derivative implies a decreasing function.

However, the absolute value in the given function stops us from differentiating right away, so we'll have to deal with it and get this function in a piecewise format.

Let's briefly consider #|x|# on its own.

On #(-oo, 0), x<0,# so #|x|=-x#

On #(0, oo), x>0,# so #|x|=x#

Thus, on #(-oo, 0), -|x|+1=-(-x)+1=x+1#

And on #(0, oo), -|x|+1=1-x#

Then, we have the piecewise function

#f(x)=x+1, x<0#

#f(x)=1-x, x>0#

Let's differentiate:

On #(-oo, 0), f'(x)=d/dx(x+1)=1>0#

On #(0, oo), f'(x)=d/dx(1-x)=-1<0#

We have a negative first derivative on the interval #(0, oo),# so the function is decreasing on #(0, oo)#

May 10, 2018

Decreasing in #(0,+oo)#

Explanation:

#f(x)=1-|x|# , #x##in##RR#

#f(x)={(1-x", "x>=0),(1+x", "x<0):}#

#lim_(xrarr0^(-))(f(x)-f(0))/(x-0)=#

#lim_(xrarr0^(-))(x+1-1)/x=1!=lim_(xrarr0^(+))(f(x)-f(0))/(x-0)=lim_(xrarr0^(+))(1-x-1)/x=-1#

#f'(x)={(-1", "x>0),(1", "x<0):}#

As a result, since #f'(x)<0# ,#x##in##(0,+oo)# #f# is decreasing in #(0,+oo)#

Graph which also helps

graph{1-|x| [-10, 10, -5, 5]}