We are asked to sketch the graph of a function subject to the following constraints:

f'(0)=f'(4)=0f'(x)=1 for x<-1f'(x)>0 for 0<x<2f'(x)<0 for -1<x<0; 2<x<4; x>4`lim_(x->2^(-1))=oo`

`lim_(x->2^(+)=-oo`

f''(x)>0 -1<x<2; 2<x<4f''(x)<0 x>4

(See attachment for sketch)

If we assume the function to be continuous we have a local...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

We are asked to sketch the graph of a function subject to the following constraints:

f'(0)=f'(4)=0

f'(x)=1 for x<-1

f'(x)>0 for 0<x<2

f'(x)<0 for -1<x<0; 2<x<4; x>4

`lim_(x->2^(-1))=oo`

`lim_(x->2^(+)=-oo`

f''(x)>0 -1<x<2; 2<x<4

f''(x)<0 x>4

(See attachment for sketch)

If we assume the function to be continuous we have a local minimum at x=-1 and a local maximum at x=2 as the derivative changes sign at a critical point. (x=-1 is a critical point as the first derivative is zero; x=2 is a critical point as the first derivative does not exist there.)**If we do not assume continuity the function could have a vertical asymptote at x=2.

At x=4 the first derivative is zero but it does not change sign on either side; the second derivative changes sign at x=4 thus there is an inflection point there.

Since the first derivative is constant on x<-2 you have a straight line with slope 1. At x=0 we have a local minimum as discussed above (also by the second derivative test.) The graph is concave up on the interval (-1,2).

At x=2 the first derivative changes sign (goes from positive to negative) while the second derivative remains positive (the graph remains concave up). Either the graph has a cusp at x=2 or the graph has a vertical asymptote at x=2 (your choice).

**Further Reading**