Question

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

Answer 1

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)`

Answer 2

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]}