Locating critical points Find the critical points of the follo...

Question

Locating critical points Find the critical points of the following functions. Assume a is a nonzero constant.

ƒ(x) = x³ / 3 - 9x

Accepted Answer

Welcome back, everyone. Determine the critical points of the function H of X equals X cubed divided by 3 minus 12 X. So whenever we want to identify the critical points, we want to first of all evaluate the derivative of the function. Let's evaluate the derivative of each of X, so we have to differentiate a 1/3 x cubed minus 12 x, and we can apply the power rule. So the derivative of X cubed is 3x2d. We are going to get X2 because 1/3 multiplied by 3 is 1. The derivative of X is once we get X2 minus 12, and now we have two cases either we set it to 0 or we check where it doesn't exist. Now this derivative exists for all X. In real numbers. Within the domain of H of X, right? H of X is a polynomial and it's defined for any X, and the derivative is also defined for any X because it's a polynomial function. So we are only going to set it equal to 0. H of X is equal to 0, and we're going to identify our critical points from these conditions. Now, if we set it equal to 0, we're going to say that X2 minus 12 is equal to 0. And now we're going to rewrite it as X2 minus square root of 122 equals 0. Because we can factor it out using the difference of squares formula. So essentially we're going to get X minus square of 12 multiplied by X + 212 equals 0. And this gives us two solutions, right, because any of the two factors can be equal to 0, so X is either positive or negative square root of 12. In simplifying, we can say that this is positive or negative, 2 of 3 because 12 is a project between 4 and 3, and we can take 4 to get that 2 and Squired of 3 as the remaining part. So we have two critical X coordinates. We want to identify the Y coordinates, and this means that first of all, we're evaluating H at the first critical point to square root of 3. Let's substitute. We're going to get 1/3. Multiplied by Now, first of all, let's cube to. And then let's cube squad of 3. We can rewrite it as squad of 3 squared multiplied by square of 3. That's the same thing as square of 3 cubed. And then we're subtracting 12 multiplied by 2 squad of 3. Now let's simplify. Well, we're going to get 1/3, multiplied by 8, multiplied by 3, square root of 3, minus 2423. Let's notice that we can cancel out 1/3 and 3, and 8 minus 24 gives us -16. So we get -16 squared of 3. And now let's identify the Y coordinate for the other critical 0.-223. We get 1/3. Multiplied by -2 cubed. Multiplied by Square root of 3 squared multiplied by a square of 3. And then we are subtracting 12 multiplied by negative 223, so we're adding 12 multiplied by 223, right? Now let's simplify. We're going to get 1/3 multiplied by -8 so we can factor out the negative sign. So 1/3, multiplied by 8, multiplied by 3, multiplied by square of 3, plus 24. Square root of 3. Once again, let's cancel out 1/3 and 3. And now we get -8 plus 24, which gives us 16. Square root of 3. And we have found the critical points. So the first one is 2. Squirrel the 3 for ax. -16 234 Y. And the second one is. -2 square root of 3. 16 square of 3. Those would be our final answers to this problem. Thank you for watching.

Locating critical points Find the critical points of the following functions. Assume a is a nonzero constant.

ƒ(x) = x³ / 3 - 9x

Key Concepts

Critical Points

Derivative

Local Extrema

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