# Trouble with local extrema graph

1. Oct 29, 2007

### coverticus

1. The problem statement, all variables and given/known data
Sketch a graph of a function f that is continuous on [1,5] and has no local maximum and minimum, but 2 and 4 are critical numbers.

2. Relevant equations

3. The attempt at a solution
Knowing 2 and 4 are critical numbers, I formed the base function x$$^{}2$$-6x+8. Not sure how to go about sketching the graph the meets the stipulations beyond this.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 29, 2007

### JasonRox

Just draw it. No need to actually have a concrete function.

3. Oct 29, 2007

### JasonRox

Also, note that x^3 has no maximum or minimum but has a critical point where? What does that point look like?

Also, are you sure it's [1,5]?

4. Oct 29, 2007

### coverticus

So just sketch a graph that has no local extrema on [1,5]? If so how are 2 and 4 critical numbers?

5. Oct 29, 2007

### coverticus

Yes it is [1,5], and x^3 has a critical point at 0, and it has a slope of zero. Correct?

6. Oct 29, 2007

### coverticus

I still need somewhat of a solid answer here, do I just sketch a graph where x=0 on [1,5] or something different? Any help here would be great.

7. Oct 29, 2007

### Dick

You sketch a graph where x=2 and x=4 have horizontal tangents, but aren't maxes or mins.

8. Oct 30, 2007

### coverticus

does that satisfy the continuity?

9. Oct 30, 2007

### Dick

Just draw the curve y=x^3 and look what happens at x=0. Now draw a curve with two points like that.

10. Oct 30, 2007

### coverticus

how can you do that without creating an extrema?

11. Oct 30, 2007

### Dick

Put your pencil on a paper at x=1. Curve up until you reach x=2 then flatten out but don't go down. Increase out of the flat part till you get to x=4, then flatten out again. Then increase some more till you get to x=5.

12. Oct 30, 2007

### coverticus

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