38. What values of a and b make f(x) = x^3 + ax^2 + bx have

Question

b. a local minimum at x = 4 and a point of inflection at x = 1?

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says determine the values of P and Q that make the function J of X, which is equal to X cubed, plus PX2 plus QX + 1 have a local maximum at X equal to -3, and a point of inflection at X equal to 2. We get 4 possible choices as our answers. For choice A, we have P is equal to -6, and Q is equal to -63. For choice B, we have P is equal to -6, and Q is equal to 63. For choice C, we have P is equal to -6, and Q is equal to -6. And for choice D, we have P is equal to 6, and Q is equal to 63. So, the first step that we need to take in order to solve this problem is determine the 1st and 2nd derivatives of our function, JFX, since we'll need to use that information along with the fact that we have a local maximum at X equal to -3 and an inflection point at X equal to 2 in order to solve for P and Q. So starting off by calculating our first derivative, we'll have J of X, and that's equal to the derivative with respect to X. Of quantity of X cubed plus p x squared. Plus QX plus 1 in quantity. And to take this derivative, we'll need to use our power rule, constant multiple rule, as well as our constant rule for derivatives. So using those, this is going to be equal to 3 x 2 plus 2PX. Plus Q We'll also need to calculate our second row, so we need to calculate J of X, and this can be equal to the derivative with respect to X. Of a first derivative, which is the quantity of 3 X 2 plus 2 PX plus Q in quantity. And so here we'll need to use our concept multiple rule as well as our concept rule for derivatives. And so using these rules, this derivative is going to be equal to 6 X plus 2P. And so now we need to determine the values of P and Q, and we'll start off by looking at the inflection point at X equal to 2. So recall in order to have an inflection point, our second derivative needs to be equal to 0 at that point. So, we need to calculate J of 2. And that's going to be equal to 6 multiplied by 2, plus 2P, and that has to be equal to 0 in order for us to have an inflection point. And so that means we'll have 12 plus 2 P equal to 0, and we need to solve this for P. So subtracting 12 from both sides, we'll have 2 P is equal to -12, and dividing both sides by 2, we'll have P is equal to minus 6. So we need to make sure that we actually have an inflection point at X equal to 2 for this value of P. So we're gonna look at points just to the left of X equal to 2 and points just to the right of X equal to 2. Let's see if our sign of our second derivative actually changes at X equal to 2. So, choosing a point to left, we'll choose X equal to 1, so we'll calculate J of 1, and that's going to be equal to 6, multiplied by 1. Plus 2 multiplied by minus 6, since we'll substitute in the value for P here into our expression. And when we evaluate this expression, this is going to be equal to -6. We also need to choose a point to the right of X equal equal to 2, so we'll choose X equal to 3. And so J of 3 is equal to 6 multiplied by 3, plus 2 multiplied by minus 6, and evaluating this expression, this is equal to 6. So to the right of X equal to 2, our second derivative is positive, and to the left of X equal to 2, our second derivative is negative. So our sign does indeed change at X equal to 2 for a second root. So we do have an inflection point when P is equal to -6. So that is the first half of our problem answered. And so now we need to determine the value of Q, and to do this, we're going to make use of the fact that we have a local maximum at X equal to -3. Now, we call that in order for us to have a local maximum at a point, that means that the first derivative has to be 0 at that point, as well as our second derivative being negative at that point. So, we'll start off by looking at the second derivative. So, we'll calculate J of minus 3. And this is going to be equal to 6, multiplied by minus 3. Plus 2 multiplied by minus 6 substituting in the value for P. and when we evaluate this expression, this is going to be equal to minus 30, which is indeed less than 0. So, now we just need to find the value of Q, that will give us a value of 0 for our first derivative, when X is equal to -3. So, we'll calculate J of -3. And this is going to be equal to 3, multiplied by the quantity of minus 3 and quantity squared. Plus 2 multiplied by minus 6. Multiplied by minus 3. Plus Q, and all that has to be equal to 0. So simplifying this expression, we will have 27 plus 36 plus Q is equal to 0. Adding together the 27 and 36, we'll get 63, plus Q is equal to 0. And now we just need to solve this expression for Q. So we just subtract 63 from both sides, and we'll have q is equal to minus 63. So this value Q will give us a local maximum at X equals to minus 3. So that is the second half of our answer here. And so, if we compare our answers here to our answer choices, we see that the correct answer choice corresponds to answer choice A. So, I hope that this explanation has been useful, and I'll see you in the next video.

38. What values of a and b make f(x) = x^3 + ax^2 + bx have
b. a local minimum at x = 4 and a point of inflection at x = 1?

Key Concepts

Local Minimum

Point of Inflection

Derivatives

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