Sketch the graph of a function that satisfies all of the given conditions:

  • f'(0) = f'(4)=0
  • f'(x) =1 if x<-1
  • f'(x)>0 if 0<x<2
  • f'(x)<0 if -1<x<0 or 2 <x <4 or x>4
  • lim x--2- f'(x) = infinity
  • lim x--2+ f'(x) = negative infinity
  • f''(x)>0 if -1<x<2 or 2<x<4
  • f''(x)<0 if x>4

Expert Answers

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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...

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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).

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