Expert: Scotto - 11/4/2008

Question
Suppose f(x) is a differentiable function for all real numbers x which also satisfies:

f '(4) = 0,
f '(x) is negative for all x less than 4,
f '(x) is positive for all x greater than 4

What can be said about the critical value x = 4?

Answer
WHen they say that f'(4) is 0, we already know an approximate value for f(x).  See, we know that the easiest way to make f'(4) = 0 is to say that f'(x) = x-4.

Now to check the second statement.  THat is, f'(x) is negative for all x less than 4.  If you put anything less than four in for x (the easiest one is 0) into the function we've got for f'(x), we get a negative answer.

Now to check the third one.  If we put a value greater than 4 into f'(x), we get a positive value, so this condition is satisfied as well.

Now I was starting to find f(x), but we don't need to do that.  Maybe I'll leave this down just to remind you that we need to look at the whole problem first.

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Since f'(4) = 0, it is known that there is an minimum, maximum, or inflection point for f(x) at 4.

When f'(x) is negative, f(x) is decreasing.
when f'(x) is positive, f(x) is increasing.

If the function is decreasing to the left of the point, has a derivative which is zero at that point, and is increasing on the other side, it is a local minimum for f(x).

Since the problem states that this is true for all x, it is an abolute minimum for the function f(x).

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Note that there are an infinite choices for the function.  To name just a few, consider f(x) = (x-4)².  Perhaps f(x) = 2(x-4)².  Perhaps f(x) = 17(x-4)^4.  Perhaps f(x) = a(x-4)^(2n) where a is positive and n is an integer greater than or equal to 1.

Perhaps something complicated, like
f(x) = (x-4)² + sin²(x) + e^[(x-4)²].

Now that's a deusy, but we're not really
in knowing the function f(x).