• Apr 30th 2007, 07:40 PM
lmao
Find the maximum and minimum values of f(x) on the given intervals.
f(x)=abs(x+1)+abs(x-1)

1>x<4 Should be greater/less than or equal to, but I dont know how to type that in.

I know how to approach normal fxns but with the absolute values I don't know. I know to just take the derivitive, set equal to zero, and solve for x, giving you the critical points. Thanks for the help!
• Apr 30th 2007, 07:58 PM
ThePerfectHacker
Quote:

Originally Posted by lmao
Find the maximum and minimum values of f(x) on the given intervals.
f(x)=abs(x+1)+abs(x-1)

1>x<4 Should be greater/less than or equal to, but I dont know how to type that in.
!

I assume you mean [1,4].

We need to find the points where f(x) is not differenciable. That is when the term inside the absolute value is zero. Thus, that happens at x=-1 and x=1. Since 1<=x<=4 we have that x=1 is the only non differenciable point. But is is an endpoint.

Now you need to take the derivative.

The derivative of y=|x| is y'=sgn(x) for x!=0 where sgn(x)=1 is x>0 and -1 is x<0.

Thus, the derivative of f(x) by chain rule is,

f'(x)=sgn(x+1)+sgn(x-1)

Note this can be zero when,

1)sgn(x+1) = +1 and sgn(x-1) = -1.

2)sgn(x+1) = -1 and sgn(x+1)=1.

If #1 then, sgn(x+1) = +1 for all x in [1,4] and sgn(x-1)=-1 never on [1,4].

If similary #2 does not ever equal to zero.

Thus, there are no critical points inside the open interval.
Just check the endpoints.