Let f(x) be a polynomial function such that f(4)=-1, f’(4)=2 and f”(4)=0. If x<4, then f”(x)<0 and if x>4, then f”(x)>0. The point (-4, -1) is which of the following for the graph of f?

A. relative maximum
B. relative minimum
C. critical value
D. inflection point
E. none of these

I'm confused on how to figure this out. Thanks for your time and help.

Since we know nothing about f'(-4) or f"(-4), we can say nothing about the properties of f there.

However, if the change in sign is inadvertent, and all the data are about x=4,

since f"(4)=0, x=4 is an inflection point.

To determine the nature of the point (-4, -1) for the graph of f, we need to consider the second derivative of f and its sign.

Since f''(x) represents the concavity of the function, f''(x) < 0 indicates a concave-downward function, while f''(x) > 0 indicates a concave-upward function.

Given that f''(4) = 0 and the given conditions:
- If x < 4, then f''(x) < 0, and
- If x > 4, then f''(x) > 0,

we can deduce that the function f is concave-upward for x > 4 and concave-downward for x < 4.

Now, let's examine the point (-4, -1).
- Since x = -4 is less than 4, f(x) is concave-downward at that point.
- Also, f(4) = -1 and f(x) = -1 at x = -4.

From this information, we can conclude that the point (-4, -1) is a relative maximum for the graph of f because the function changes concavity from downward to upward at that point.

Therefore, the correct answer is:
A. relative maximum.

To determine the nature of the point (-4, -1) on the graph of the function f(x), we need to consider the information about the first and second derivatives of f(x) at x=4.

Given that f'(4) = 2, we can conclude that the function has a positive slope at x = 4. This suggests that the graph of f(x) is increasing (as it moves to the right) near x = 4.

Now, considering f''(4) = 0, it tells us that at x = 4, the graph of f(x) has an inflection point. An inflection point is a point on the graph where the curvature changes. At an inflection point, the concavity of the curve can switch from being concave up to concave down, or vice versa.

Since we know that f''(x) changes sign at x = 4 (i.e., f''(x) < 0 for x < 4 and f''(x) > 0 for x > 4), it implies that the graph of f(x) changes concavity at x = 4. Precisely, f(x) changes from being concave down before x = 4 to concave up after x = 4 (or vice versa).

Now, let's relate this information to the point (-4, -1). Since (-4, -1) is to the left of x = 4, it falls within the interval where f''(x) < 0. Therefore, the graph of f(x) is concave down to the left of x = 4. In terms of the y-coordinate, a concave down graph has a local maximum.

Hence, the correct answer is:

A. relative maximum.

The point (-4, -1) is a relative maximum for the graph of f(x).