119. Find the values of constants a, b, and c such that the gr...

Question

119. Find the values of constants a, b, and c such that the graph of y = ax^3 + bx^2 + cx has a

local maximum at x = 3, local minimum at x =- 1, and inflection point at (1, 11).

Accepted Answer

Hello. In this video, we are going to be solving for the constants M, N, and O, such that the cubic function Y is equal to MX cubed plus NX2 plus OX exhibits a local maximum at X is equal to 3, a local minimum X is equal to 1, and an inflection point at 28. So the idea for this problem is that we want to create a system of equations in order to solve for these coefficients. Now, if we take a look at the last part of this question, we are told that we have an inflection point located at 2-8. What this means is that if we input 2 as X into our original equation, the output of the equation should be -8. So, let's go ahead and plug in this point into the given cubic equation. That is going to give us M multiplied by 2 cubed, plus N multiplied by 2 squared, plus O multiplied by 2 is equal to -8. Simplifying this equation will give us 8 M + 4N + 20 equal to -8. We will call this equation one, and we will come back to it later. No We are also told that this function has a local maximum. And a local minimum located at the values X is equal to 3 and X is equal to 1. In order to obtain these values, we need the second, we need the first derivative of the original function. So, let's go ahead and take the first derivative of the given cubic function. That will give us Y is equal to 3 MX 2 plus 2NX plus O. Now, in order to obtain the local maximum and minimum, we want to find the values of X that cause this derivative to be zero. Now, since we know that the locations of the local extrema are at X is equal to 3 and X is equal to 1, we can create two more equations by plugging in these values of X into the derivative and setting the derivative equal to 0. If we were to plug in one, this will give us 3M + 2N plus O equal to 0. We will call this equation 2 and come back to this later. Now, plugging in 3 to the to the derivative is going to give us 3M multiplied by 3 squared, plus 2N multiplied by 3 plus O equal to 0. This will simplify to be 27 M plus 6N plus O equal to 0, and this will be our third equation. Notice that we have 3 equations to help us solve for 3 variables. Now we can go ahead and treat these 3 equations using a system of equations to solve for M, N, and O. To start this problem, let's go ahead and pair up the 1st and 2nd equation together. That is going to give us 8 M + 4N + 20 equal to -8, and we are going to add the second equation, which is 3M + 2N plus O equal to 0. What we want to do now is eliminate some of the parameters. We can eliminate N by multiplying the second equation by -2. This is going to leave us with 8 M + 4N + 20 equal to -8, and we will be adding the equation -6M minus 4N minus 20 equal to 0. When we add these equations together, we end up canceling two parameters, and in O, and we are left with 2 M equal to -8. Dividing two to both sides of this equation will leave us with M is equal to -4, which is going to be the first parameter that we are looking for. Now what we want to do is we want to create another equation in terms of M and one of the other parameters, and use this M to solve for it. Let's go ahead and pair up equation 2 and equation 3 together. That is going to give us 3M plus 2N plus O equal to 0 and 27M plus 6N plus O equal to 0. Now, what we want to do is you want to go ahead and eliminate either N or O from this equation. The easiest one to eliminate here is going to be O, so let's just go ahead and multiply this first equation by -1. That will give us -3M minus 2N minus O equal to 0, and we will be adding 27M. Plus 6 N + 0 equal to 0. Now when we add the Os together, these are going to eliminate these parameters. Now what we need to do is we just need to sum up the rest. 27 m minus 3M will leave us to a 24M. And 6N minus 2N will leave us with positive 4 N and this will equal to 0. Now what we can do is we can plug in our M to solve for N. This will be 24 multiplied by -4 plus 4 N equal to 0. 24 multiplied by -4 will give us -96, and if we add that to the right of the equation, we have 4 N is equal to 96. Finally, if we divide 4 to both sides of this equation, we get N is equal to 24, which is going to be the second parameter we are looking for. Now, in order to solve for O, we can take our value of M and N and plug it into any of the three equations that we came up with. Let's go ahead and plug them into equation 2. That is going to give us 3M, which is -4, plus 2N which we just solved to be 24. Plus 0 is equal to 0. 3 multiplied by -4 will give us -12, and 2 multiplied by 24 will give us 48. Now -12. Plus 48 is going to give us 36, and if we subtract that to the right hand side, we get O is equal to -36, which is going to be the final parameter that we need to solve for. So to conclude, by creating a system of equations, we found that the three coefficients are going to be M is equal to 4, N is equal to 24, and O is equal to -36. So I hope this video helps you in understanding how to approach this problem, and I will go ahead and see you all in the next video.

119. Find the values of constants a, b, and c such that the graph of y = ax^3 + bx^2 + cx has a
local maximum at x = 3, local minimum at x =- 1, and inflection point at (1, 11).

Key Concepts

Critical Points and Derivatives

Second Derivative Test

Inflection Points

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