Which figure is the problem referring to?
Some general statements that might help while we don't have the figure:
a) The x-coordinates at which f has critical values are every c in [-3, 4] such that f' is either zero or undefined at c. f has a relative maximum at c in (-3, 4) if there is a neighborhood around c such that f'(x)>0 left of c and f'(x)<0 for all x in that neighborhood; f has a relative minimum at c in (-3, 4) if there is a neighborhood around c such that f'(x)<0 left of c and f'(x)>0 right of c in that neighborhood.
(In more symbolic notation, this would read, f attains a relative minimum at c if there exists a δ>0 such that f'(x)<0 for all x in (c-δ, c) and f'(x)>0 for all x in (c, c+δ); f attains a relative maximum at c if there exists a δ>0 such that f'(x)>0 for all x in (c-δ, c) and f'(x)<0 for all x in (c, c+δ)).
Often, one can find this neighborhood by observing the sign changes of f' at the critical points of f. For example, if you have f'(x) = (x-a)(x-b)^2(x-c) with a<b<c, observe that there is no sign change at b, where as there are sign changes at a and c. If x>c, then f' is positive; if x<c, then f' is negative; this means that f attains a local maximum at c.
Also check the boundary points -3 and 4 to see if they are maximums and minimums (that is, if they are critical points)
b) f is nonincreasing on an interval I if f'(x)≤0 for x in I; x is strictly decreasing on an interval I if f'(x)<0 for x in I. (I'm not sure if the problem means nondecreasing or strictly increasing when it says "decreasing"-- I think it usually means "nonincreasing" when "strictly" is ommitted). f is concave up on an interval I if f''(x)>0 for x on I.
c) The inflection points of f are c on [-3, 4] are where f''(c)=0 and the sign of f' changes at c (i.e. where f changes from concave up to concave down or vice-versa).
I hope this helps!
