Given: 𝑓(𝑥) = 𝑎𝑥^3 + 𝑏𝑥^2 − 3 and 𝑓"(−2) = 0. If it is further given that the point (−3; 6) lies on the graph of 𝑓. Show that 𝑎 = 1/3 and 𝑏 = 2.
We start off by finding the first derivative of equation f(x) = ax3 + bx2 - 3: f'(x) = 3ax2 + 2bx. We now take the second derivative of equation f, because we have been told that f"(-2) = 0: f"(x) = 6ax + 2b (1). We know that with an x value of -2, equation (1) is equal to 0: f"(-2) = 6a(-2) + 2b = 0-12a +2b = 0 (2). This equation will be used later to find the final answer. We also know that the point (−3; 6) lies on the graph of 𝑓. Therefore, for an x value of -3, f(x) equals 6: f(-3) = a(-3)2 + b(-3)2 - 3 = 6-27a + 9b = 9 (3). We then solve equations (2) and (3) simultaneously, as we have two unknowns and two equations, and reach the following answer: a = 1/3 ; b = 2. This question would be worth a total of 6 points.
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