Let ƒ(x) = (x - 3) (x + 3)²

e. Identify the l...

Question

Let ƒ(x) = (x - 3) (x + 3)²

e. Identify the local extreme values and inflection points of ƒ .

Accepted Answer

Welcome back everyone to another video. Consider the function G of X equals x + 2 multiplied by x minus 4 squad. Determine the local Xtrema and points of inflection for G of X. So first of all, let's identify the first derivative of the function, and because we have a product, well, we are going to apply the product rule which says that first of all we're taking the derivative of The first factor multiplying by the second factor, which in this case is x minus 42, and then we're adding the first factor X + 2, which is multiplied by the derivative of the second factor, so X minus 42. And now we want to evaluate those derivatives. So the derivative of X + 2 is 1. This means that we end up with X minus 42. And then we are adding x + 2. Which is multiplied by the derivative of X minus 4 squad, and it is a composite function. So first of all, we apply the power rule which gives us 2 multiplied by x minus 4. The original exponent decreases by 1 unit, and then according to the chain rule, we're multiplying by the derivative of X minus 4, which is simply 1. So now what we can do is simplify. And also let's factor out X minus 4, which is a common factor. So we get X minus 4 and C is x minus 4 for the first term, and now for the second term we have + 2 in X + 2. Right? We can also cross out the derivative of X minus 4 because it is equal to 1, and now we are going to simplify it, so we're going to get the X minus 4 multiplied by. X minus 4 + 2 x + 4. So, let's combine the like terms. We get 2 X plus X, which gives us 3 X, and then minus 4 + 4, right? So this gives us 0. Overall, the derivative is simply We Multiplied by x minus 4. And now what we are going to do is simply identify the critical points, right? Notice that our derivative exists for any X in the domain of G of X, which is also all real numbers, so we're just going to set it equal to 0, which means that either our first factor is equal to 0. This gives us The first critical X coordinate X1 equals 0, or X minus 4 is equal to 0, which means that our second critical X is 4. And now what we want to do is simply identify the second derivative. To classify these as local minimum or local maximum. So to find the second derivative, we're going to distribute. The first derivative, that's 3 X2 minus 12 X, and now applying the power rule, we're going to get 6 X minus 12. Now if we consider the second derivative at 0. Substituting 0 for X, we're going to get negative 12, this is negative, and if it's negative, it means that the function is concave down, so we have a point of maximum. So X equals 0 is a local maximum. And now let's identify the second derivative at 0.4, substituting 4, we get 24 minus 12, that's 12, and this value is positive, meaning the function is concave up, and we have a point of minimum. Now we're going to identify. The second derivative, which we already have is 6 x minus 12, and specifically what we want to determine are the inflection points. So we're going to set it equal to 0, which means that 6 X minus 12 is equal to 0 and X is equal to 2. And essentially we want to check if the second derivative changes sign. So for example, if we evaluate G at any point that is lower than 2, such as G at 0, we can notice that it's negative. And now we want to choose any point that is greater than 2, such as G at. 4. We already know that it's positive. Notice that we have a change of sign, meaning X equals 2 is an inflection point. And that's it. Now we want to identify the Y coordinates, so we are going to the original function and evaluating G at.0, which gives us 0 + 2. That's simply 2 multiplied by 0 minus 42, so -42 is 16, and this gives us 32. Then we want to evaluate G at 0.4 because we have 4 minus 4 as one of our factors. Well, this gives us 0. And finally, we want to evaluate G at the inflection point. So G at 0.2 is 2 + 2, which is 4, multiplied by 2 minus 42. So negative 2 squad is 4, and we get 16. So now we can classify them. We have a local. Maximum Which is 0 for the X coordinate. 32, that's our Y coordinate. We also have a local minimum. Notice that it's at X equals 4 and Y of 0. And then we have our inflection point. It's at X equals 2 and Y of 16. Well then, we have identified the final answers to this problem. Thank you for watching.

Let ƒ(x) = (x - 3) (x + 3)²

e. Identify the local extreme values and inflection points of ƒ .

Key Concepts

Local Extreme Values

Inflection Points

First and Second Derivatives

Watch next