<p>Ok, here it is:</p>
<p>for what values of a and b does the function</p>
<p>f(x) = x^3 + ax^2 + bx + 2</p>
<h2>have a local maximum when x= -3 and a local minimum when x= -1?</h2>
<p>The teacher went over this problem in class and I remember him solving a system of equations, but I still can't get it. </p>
<p>If anyone could provide, step by step or a good explanations, I would highly appreciate. Thank you.</p>
<p>I'm not a genius... but I'll try</p>
<p>local max or min is when the slope = 0, so you find where f'(x) = 0.</p>
<p>f'(x) = 3x^2 + 2ax + b</p>
<p>3x^2 + 2ax + b = 0</p>
<p>Substitute in the x=-3 and x=-1 in the derivative equation, since you know those are the values when f'(x) = 0.</p>
<p>27 - 6a + b = 0
3 + 2a + b = 0</p>
<p>Subtract second eq from first and get:</p>
<p>24 - 8a = 0</p>
<p>a = 3
b = -9</p>
<p>And there you are. But be wary, frasifrasi, as I'm no genius.</p>
<p>If you need any more explanation, just ask.</p>
<p>thanks lin, that was quick! One more thing, aren't those candidates max and mins? don't we have to take the second derivative and verify them using concavity or something? In other words, take f" of those numbers and if it is distinctively pos or neg, then we have a hit. If it is a 0, we don't--hopefully I am not confusing you.</p>
<p>Thanks once again.</p>
<p>ordinarily when you aren't given the specific type of peak/valley (max/min), you need to verify with the second derivative test...if it is negative, the original function is concave down (looks like a frown--> n) and so it's a max...if it's positive, the original is concave up (looks like a cup--> u) and so it's a min...</p>
<p>hope this helps!</p>
<p>Well, if you wanted to find the minimums from the equation, you would have to either use the second derivative and then if the x value is negative when substituted into the second derivative eq, it's a max, and if it's positive, it's a min, or just find whether the first derivative is going from positive to negative or vice-versa at the point you're interested in.</p>
<p>But since the problem tells you that that's where the min and max are, you don't need to concern yourself with that. You just assume they are what the problem says they are.</p>
<p>Hope that answers your question.</p>
<p>Yes, of course... : )</p>
<p>iin77, you made a small mistake. The system of equations you need to solve is </p>
<p>27 - 6a + b = 0
3 - 2a + b = 0</p>
<p>giving a = 6 and b = 9 as the solution.</p>
<p>oh =-) sorry, lol</p>
<p>Must be because I'm not a genius ;-)</p>