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 + 𝓍² ― 8𝓍 + 5

Accepted Answer

Welcome back everyone to another video. For the function Y equals x cubed minus 3 x 2 minus 9 x + 5, find the extreme values absolute and local over its natural domain, and determine where they occur. First of all, let's identify the domain, because we're given a polynomial. This means that X belongs to all real numbers, right? We don't have any restrictions. Now we can identify the extreme values by differentiating the function. So differentiate X cubed minus 3 X2 minus 9 X + 5. We can apply the power rule which gives us. 3 X2 minus 6 x minus 9. And now we want to set it equals to 0. So, first of all, let's factor out 3. In C we get X2 minus 2X minus 3. We set it equals to 0. Dividing both sides by 3, we get X2 minus 2X minus 3 is equal to 0. And now we can factor out the quadratic term. Which gives us X minus 3, multiplied by X + 1, and we set it equal to 0. Solving this equation, we set each factor equal to 0. Either X minus V is 0. 4 X + 1 is 0. From the first equation, we can see that X equals 3. From the second one, X equals -1. So we have identified the critical points. And then we're going to classify them using the second derivative test. So let's identify the second derivative, Y is equal to, we can factor out 3 and differentiate X2 minus 2X minus 3. Which gives us 3 multiplied by 2 X minus 2. And now let's evaluate the second derivative at each critical point. So why at 3? The sequel to we multiplied by. 2 multiplied by 3 minus 2. Which is 6 minus 2, that's 4, and 4 multiplied by 3 is 12, so that's positive. Now if the second derivative. Gives us a positive value. This means that the function is concave up. And therefore we have a local minimum. Similarly, we are going to evaluate Y at -1. Which is we multiplied by 2 multiplied by -1 minus 2. That's -2 minus 2, which is -4, multiplied by 3, that's -12, and that's less than 0. The function is concave down. Which indicates a maximum. Well done, so we can state that there is a local. Minimum At X equals 3. And now we have to identify the Y coordinate. We can do that by evaluating Y at 3. Which is 3 cubed. Minus 3 multiplied by 3 squad, minus 9 multiplied by 3. Plus 5 This gives us -22. So we have a local minimum at 3.22. Then we have a local maximum. At -1. We can identify the Y coordinate by finding Y at -1. Which is -1 cubed. Minus 3 multiplied by -1 squad. Minus 9 multiplied by -1 + 5. And this gives us 10, so we have a local maximum at -1.10. And now what about the absolute extreme values? Well, we have a cubic polynomial with. A positive leading coefficient of one, which means that. If X approaches an affinity. Then why grows infinitely large as well. If x approaches negative infinity. Why approaches negative infinity as well. This means that we have no. Absolute Extreme values. And that's it. We only have local minimum and local maximum values. 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 + 𝓍² ― 8𝓍 + 5

Key Concepts

Critical Points

First Derivative Test

Second Derivative Test

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