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Let f(x) be a polynomial function such that f(-2)=5, f'(-2)=0 and f"(-2)=3. The point (-2,5) is which of the following for the graph of f?

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2 Answers

I completely agree with George's method, and it does work for this case because it was stated that this is a polynomial function. But for more general cases, here's how you would approach this problem. 

The first thing we look at is the relationship between the original function and it's derivative, ƒ'. The relationship is as follows:

1. ƒ' is negative -> ƒ is decreasing. 

2. ƒ' is positive -> ƒ is increasing.

3. ƒ' = 0 -> ƒ has a critical point. Either rel. max, rel. min, or some inflection points. 

The second thing we look at is the relationship between the original function and it's second derivative, ƒ". This is their relationship:

1. ƒ" is positive -> ƒ is concave up.

2. ƒ" is negative -> ƒ is concave down.

3. ƒ" = 0 -> ƒ may have an inflection point given that ƒ" has different signs to the left and right of the point being examined.

For our problem here we can apply these rules quite easily. Since we're examining the same point in both relationships, x=-2, we'll leave it out of the work for the sake of having less clutter. 

The relationship between ƒ and ƒ' it seems is the third on the list where ƒ has either a rel. max, rel. min, or an inflection point, because here we find ƒ' to be 0. To be exactly sure of which it is, we look at our other criterea using ƒ". This relationship between ƒ and ƒ" is the second because ƒ" is 3, a positive number, which means that ƒ is concave up, much like a U. So far we have eliminated the last three anwers based on our criterea and now we must do some thinking about whether this is a rel. max or rel. min. Well, if we think of something being concave up, it must mean that to the left of the point we have the function coming down and to the right, it's going up, that's how we end up with the U looking shape. Having said that, the answer must be rel. min because the only place for the function to go down to and up from is the min. So our answer choice is in fact (b), which agrees perfectly with George. 

Hope this helps! 

f" = 3

f' = 3x + c1

f'(-2) = 0 = 3(-2) + c1, c1 = 6,

f' = 3x + 6

f = (3/2)x^2 + 6x + c2

f(-2) = 5 = (3/2)(-2)^2 + 6(-2) + c2, c2 = 11

f = (3/2)x^2 + 6x + 11.

A critical point occurs at x = -2,  there is no inflection as f" is always (+) and so a relative minimum.

Foil (b) is the correct answer.