Lois C. answered 06/20/20
BA in secondary math ed with 20+ years of classroom experience
We'll start by taking the first derivative of the function which is f '(x) = 3x2 + 6x - 9. Factoring out the GCF of 3, then factoring the remaining trinomial, we have 3( x + 3) (x - 1 ). Setting this equal to 0, we have our critical points occurring at x = -3 and x = 1.
Now what we can do is partition the number line into 3 sections, using our critical points as the dividers of the number line, and check random values of x from each partition into the rule for f '(x) and see whether the slope/derivative at those x values is positive or negative. So the first section of the number line would be the values ( -∞, -3 ), so we could select, say, x = -5 from that part of the number line. Inserting it into the rule for f '(x), we get a value of 3( -5)2 + 6(-5) - 9 or 36. From the next section of the number line, between -3 and 1, we could use an easy value of x at x = 0, and f '(0) is -9. From the final section of the number line, between 1 and ∞, we could use x = 2 and f '(2) = 3(2)2 + 6(2) - 9 or 15.
Now for the maximums and the minimums: since the slope changes from positive to negative at -3, we would have a max at x = -3. Since the slope changes from negative to positive at x = 1, we know that we have a minimum at x = 1.
Graphing the original function on the graphing calculator should confirm these max and min locations.