Find values of a and b such that the function

Question

ƒ(𝓍) = (a𝓍 + b) / 𝓍² ―1)

has a local extreme value of 1 at 𝓍 = 3. Is this extreme value a local maximum or a local minimum? Give reasons for your answer.

Accepted Answer

Hello, in this video, we are going to be solving for the values A and B that cause the given function to have local extreme value of 2, and X is equal to 3. Then we want to determine if this is a local maximum or minimum value. So, the idea of this problem is that we're going to create a system of equations of two variables in order to solve for A and B. Now the problem tells us that the function has a local extreme value of 2, and X is equal to 3. What this means for the original function is that when we plug in the value of 3, we get the output of 2. So what we are going to do is we are going to create our first equation by plugging in 3.2 into the original function. The original function is F of X equal to A X plus B divided by X2 minus 4. And by plugging in these these values, we get A multiplied by 3 plus B divided by 3 squad minus 4 is equal to 2. This will simplify to leave us with 3 A plus B, divided by 5 is equal to 2, which will further simplify to be 3A plus B equal to 10. This is going to be our first equation for A and B, so we will set it to the site for now and come back to it later. Now, we have this local extreme value at X is equal to 3, but in order to obtain the local extreme value, we need the derivative of the function. So by taking the derivative with respect to X, and by using the quotient rule, we get the function F of X is equal to negative AX2 minus 2BX. Minus 4A. is divided by the quantity. X squared -4 quantity squared. Now, in order to obtain the critical points here, we need to find the values of X that caused the numerator to be 0. So that will give us negative AX2 minus 2 BX minus 4A is equal to 0. Now, again, we know that the extreme value is located at X is equal to 3. So if we plug in X's equal to 3 into this equation, we will get our second equation for A and B. This will leave us with negative A multiplied by 32, minus 2B multiplied by 3, minus 4A is equal to 0. This will simplify to leave us with -9A. -6B minus 4A is equal to 0. And this will further simplify to be negative 13A minus 6B is equal to 0. This is going to be our 2nd equation for A and B. And now that we have two equations of A and B, we can go ahead and solve for them by using a system of equations. Let's go ahead and take the first equation and add it to the second one. So we will add 3 A plus B equal to 10. Now, we will go ahead and eliminate one of either A or B. Let's go ahead and eliminate B from the equations by multiplying the second equation by 6. This will give us -13A, minus 6B is equal to 0, and this will give us 18A plus 6B is equal to 60. By adding the equations together, we eliminate B. And this will give us 5A is equal to 60, which will further simplify to be A is equal to 12. So, this is going to be the first variable that we are solving for. Now that we have the value of A, we can plug this value of A into the 1st 1st equation, and this will allow us to solve for B. That is going to give us 3 multiplied by 12, plus B is equal to 10. This will simplified to be 63 plus B is equal to 10, and that will give us B. I, oh sorry, 3 times 12 is 36, not 63. But this will give us 36 plus B equal to 10, and that is going to simplify to be negative 26, which is the second value for B. Now that we've found the variables A and B, we want to verify that X is equal to 3, is a local maximum or minimum value. In order to do this, we will use the first derivative test. We will create a number line and place the value of 3 here. And what we are going to do is we are going to take testing points to the left and right of 3 to plug in the first derivative. We can plug in 2.5 and 4. If we plug in 2.5 into the derivative F of 2.5 gives us an approximate value of 1.38. But since this is positive, the graph will be increasing in this region. And if we plug in 4 into this first derivative, F4 will give us the value of negative 0.22. This is negative, so the graph is decreasing. So as we're passing through the value 3, the graph switches from increasing to decreasing, which verifies that we have a local maximum. At the value X is equal to 3, and this is going to be the final solution to the problem. 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.

Find values of a and b such that the function

ƒ(𝓍) = (a𝓍 + b) / 𝓍² ―1)

has a local extreme value of 1 at 𝓍 = 3. Is this extreme value a local maximum or a local minimum? Give reasons for your answer.

Key Concepts

Derivative and Critical Points

Second Derivative Test

Function Evaluation

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