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Problem 73 Hard Difficulty

Find a cubic function $ f(x) = ax^3 + bx^2 + cx + d $ that has a local maximum value of $ 3 $ at $ x = -2 $ and a local minimum value of $ 0 $ at $ x = 1 $.

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Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

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How Derivatives Affect the Shape of a Graph

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Video Transcript

All right, So we're being told to find a cubic function f of x equals X cubed, possibly x squared. Plus the 8. 50 that has a local max of three at X equals negative two in a local men of 0 64 1. So that's all the information we're given. So the first thing we want to do is take the derivative. So we want to take the derivative of primer back, and this will give us three a x squared plus b not B expert going to be to be xr to be x plus c. And so, um, the reason I took the derivative is because we're told some information about the derivative. Believe it or not, we're told that the local Max, um, and a local men So what is the local Max and local men means that there has to be a point at local maximum ensures that F prime is equal to zero. So we're told that at these values, negative two and one, So I have prime of negative. Coup is equal to f prime of one because that's when we have a local max. So when a local massacres, when it goes like this or in the local ministers like this. But both have a derivative zero. So we know that this fact is true. Mm. So what do we do now? We Well, we can just plug in our values into X for F prime. So we get three a, then you get negative. Two squared. No, no. Um, no, no, I'm sorry. This is going to be, uh So what? We're going? Yeah. So we're going to plug in. We're gonna plug these numbers into, uh, X right here. So we're gonna get three a and the negative two squared, plus, um, to be and then negative too. And then plus C and then we're gonna plug in one. So for one, we just get three a plus two b plus c. And then once you simplify this, you get 12 a minus four b equals three. A push to be because the sea is canceled. And now we can solve for either air B. I solve for B in this case. So he's all for B. You get three have. Hey, so now we what we do as we, um we know that f prime of one. So I'm gonna go ahead and put in red. So this is our second step. So we know that f prime of one is equal to zero. Yeah. And so what we're gonna do is we're gonna set, um three a X squared, just to be x plus C equals zero. So equal zero. And we're gonna plug in one. So this is going to come out to be, um, zero. It's equal to three. A plus to be. Let's see. Well, I'm gonna bring this on to the next page. I'm gonna rewrite this so it's gonna be zero. Is he called. I'm gonna go ahead and continue the red color to show are different. Step Oops, sorry. Zero is equal to three. Um, a plus to be plus c now, reminder. We saw for B so we can actually plug in or Newbie. So zero goes three a plus, two times, 3/2 a and m plus c. And now what we can do is we can just all for C galaxy is gonna give us negative six A. So now I'm gonna circle some of the important, um, constant. So we have b go to three half a and see you to go the negative six a. Now we're gonna move on to another step, and that is f of negative two. So what we're going to do so have a negative too. Now that we know the values of, um hey, we can actually go ahead and start plugging this into our original function. I mean, we know the value of B and C, so we're gonna go ahead and start plugging this into our original function. Yeah, so we're going to get f of negative two. So that's gonna give us negative tube cube times a and then plus negative two squared and in times be would be is three half a and then we're gonna get Plus, this is gonna be negative two times, see, and then plus D. And this is equal to three because this is told to us effort at X equals negative two. We get three mhm. We're going to hold on to this value right here and now we're going to solve for f of one so half of one. Well, that's going to be a plus three half a, which is our B. But now it's three half a rewritten then plus C and then plus D. And now this can be rewritten as five half a, um, plus the plus d and this is all equal to zero. Then we also know that f of negative two. You can also, um, So what we can also do now is we're going to do, um, f of negative two minus f of one. And that's gonna give us three minus zero, which is still equal to three. So f of two minus f uh, one. This is delicate three. And then this is equal to negative to a minus five. Half a managed to see minus c. So I did some substitution here. Um, Now we can get another value for See that we can rewrite to see negative one minus three, half A. We also know. So now we have to seize. Now that we have to seize, we cannot just solve for what number A is. Remember, we succeed equal to negative six A. So now we have two versions of C, so we can say negative. Six a is equal to negative one minus three half. Okay. And if you saw for this, you get a you go to to ninth. Wow. So Oh, that's great. Um, now we know a So now that we know A we also know that B is equal to three half times a so three half and then we know we just saw for a is 29 to 9. So you get 39 which is just one third. So now we don't be well, now that we know what B is and we know what a is Remember, let's see what's equal to negative six a. And we know what a s a negative six time to ninth and that comes out to be negative for third. So now we have a B A B N c. And to find what D is we can There's a There's a number of ways to do this. I'm just gonna do the simple one. So everyone which is a B plus a plus B plus C plus d, which is gonna be to ninth plus one third and then she is negative. So it's me minus four third and then plus D and we know that everyone is equal to zero. And if you saw for D, you got 7/9. So now we have salt for A B and A, B, C and D. And if you plug in those values into our original function, I mean to our original form, that is the cubic function.

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For what values of the numbers $ a $ and $ b $ do…

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Calculus: Early Transcendentals

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Anna Marie Vagnozzi

Caleb Elmore

Kristen Karbon

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Anna Marie Vagnozzi

Caleb Elmore

Kristen Karbon

James StewartCalculus: Early Transcendentals

Grace He

Anna Marie Vagnozzi

Caleb Elmore

Kristen Karbon

James Stewart

Calculus: Early Transcendentals