Finding Extreme Values
In Exercises 1–1...

Question

Finding Extreme Values

In Exercises 1–10, find the extreme values (absolute and local) of the function over its natural domain, and where they occur.

y = 𝓍³ ― 2𝓍 + 4

Accepted Answer

Welcome back, everyone. For the function Y equals x cubed minus 3X2 + 1, find the extreme values, absolute and local over its natural domain and determine where they occur. So we're given our function Y equals x cubed minus 3 x2 + 1. And first of all, let's identify the domain. The domain is all real numbers because we're given a polynomial expression. What we're going to do is simply begin by identifying the local extreme values. And we can do that by identifying the first derivative, which is Y. So let's differentiate X cued minus 3 X2 + 1. According to the power rule, the derivative is 3x2 minus 6 X. And we want to identify the critical points by setting it equal to 0. We can factor out 3 Xs which leaves us with X minus 2 inferences. We're setting it equal to 0. And we can solve this equation by setting each factor equals to 0. So either we axis equals to 0, which gives us X equals 0 as our First, critical point or X is equal to 2, considering the second factor. Now that we have 2 critical points, we have to classify them as a minimum or a maximum. So what we're going to do is apply the 2nd derivative test. So it's differenti we. X squad. Minus 6 X. Once again, we can use the power rule, which gives us 6 X minus 6. Or baseless 6 multiplied by X minus 1. And let's evaluate the 2nd derivative at each critical point. So why add 0 is equal to 6 multiplied by 0 minus 1. Which is -6. And then Y double prime at 2. Is equal to 6, multiplied by 2 minus 1. Which is 6. Now, -6 is less than 0. Which means that our function is concave down. And if it's concave down, it indicates a local maximum. And 6 is greater than 0, which means that the function is concave up. And according to the 2nd derivative test, this indicates a minimum. So now let's identify the Y coordinates for the maximum and the minimum by evaluating Y at each point. So Y of 0. Is equal to 0 cub. Minus 3 multiplied by 02 + 1. And that is equal to one. And Y at 2 is equal to 2Q. Minus 3 multiplied by 22 + 1. Which is equal to -3. So we can conclude that we have a. Local Minimum At X of 2 and Y of -3. We also have a. Local maximum. At X equals 0. And Y of 1. And now let's consider the absolute extreme values. Let's notice that we have a cubic polynomial with a positive leading coefficient of one. Which means that as X approaches infinity, Y approaches infinity as well. If X approaches negative infinity. Why approaches negative infinity as well. This means that we have no. Absolute Extreme values. Well done, so we only have local extreme values and no absolute extreme values. That would be our final answer to this problem. Thank you for watching.

Finding Extreme Values
In Exercises 1–10, find the extreme values (absolute and local) of the function over its natural domain, and where they occur.

y = 𝓍³ ― 2𝓍 + 4

Key Concepts

Critical Points

First Derivative Test

Second Derivative Test

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