The Function f(x) from the graph f'(x)

1. Since the function increases when the slope is positive, or when f'(x)>0, this function is increasing in the interval (-2,0)U(0,2). Here, the slope is positive, since the outputs of the graph f'(x) are positive. Thus, the function is constantly increasing. Since a function is decreasing when the slope is negative, or when f'(x)<0, this function is decreasing in the interval (-infinity, -2)U(2, infinity). Here, the slope is negative, since the outputs of the graph f'(x) are negative. Therefore, the graph is constantly decreasing.

2. There is a relative extrema at the points (-2,0) and (2,0). Here, the slopes changes from positive to negative or negative to positive.
At the point (-2,0), slope changes from negative to positive. This is a local/relative minimum.
At the point (2,0), slope changes from positive to negative. This is a local/relative maximum.

3. The function is concave up when f"(x)>0 and concave down when f"(x)<0. In order to find where the graph of f(x) is concave up or concave down, one must take the derivative of the function f'(x), or, in other words, find its slope.
f(x) is concave up in the interval (-infinity, -1.2)U(1.2, infinity). Here, f'(x)>0 (above the x-axis).
f(x) is concave down in the interval (-1.2,0)U(0,1.2). Here, f'(x)<0 (below the x-axis).

4. Since the graph of f'(x) has 4 "bumps" (i don't know how else to explain it), this means that f(x), from which f'(x) is derived from, must be one power higher. Therefore, the graph of f(x) must be an x^5 function. Also, from using my data from f'(x) and f"(x), i was able to draw the graph of f(x). f'(x) provided me with the slope at certain intervals, and f"(x) provided me with the concavity at certain intervals. This helped me finally conclude that the graph of f(x) is indeed an x^5 function.