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!

2. 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.