Verify that the following functions satisfy the conditions of ...

Question

Verify that the following functions satisfy the conditions of Theorem 4.9 on their domains. Then find the location and value of the absolute extrema guaranteed by the theorem.

f(x) = x√(3-x)

Accepted Answer

Hello there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Check if the function F of X is equal to 2 X multiplied by the square root of 6 minus X satisfies the conditions of the following theorem on its domain. If it does, identify the location and the value of the absolute extremum guaranteed by the theorem. Theorem. Suppose F is continuous on an interval I that contains exactly 1 local extremum at C. If the local maximum occurs at C, then F C is the absolute maximum of F on I. If a local minimum occurs at C, then F of C is the absolute minimum of F on I. Awesome. So it appears for this particular problem, we're asked to check if this function F of X satisfies the conditions set to us by the following theorem that's provided to us within its domain. And if it does lie on its domain, then we need to identify the location and value of the absolute extrem guaranteed by this specific theorem. So with that said, now that we know that we're what we're ultimately trying to solve for, let's read off our multiple choice answers to see what our final answer might be. A is absolute maximum of 4 multiplied by the square rooted 2 at X equals 2. B is the absolute minimum of 8 multiplied by the square rooted 2 at X equals 4. C is the absolute maximum of 8 multiplied by the square rooted 2 at X equals 4. And D is the theorem does not apply to the function. Fantastic. So our first step is we need to identify the domain. So we need to know to recall that the function. F of X is equal to 2 X multiplied by the square root of 6 minus X involves a square root, and the expression inside of the square root, the 6 minus X, must be non-negative in order for F of X to be a real value. So therefore, the domain as a result, has to be 6 minus X is greater than or equal to 0, so that means when we rearrange this equation to isolate and solve for X, that means that X has to be less than or equal to 6. So therefore, the domain of F of X is X is less than or equal to 6. For our second step is we need to check the continuity. So the function F of X is equal to 2 X multiplied by the square root of 6 minus X will be continuous on its domains negative infinity 62. Where a parenthesis indicates an open interval, whereas a square bracket indicates a closed interval. And it will be continuous on its domain negative infinity or parenthesis negative infinity 62 bracket because it will be a combination of continuous functions, meaning a linear function and a square root function. So with that said, Our next step or step 3, is we need to find, or rather determine the critical points. So as we should recall, in order to find the critical points, we must first compute or determine the Derivative of F of X and set it equal to 0. So we need to take F of X is equal to 2 X multiplied by the square root of 6 minus X, and we need to find the derivative of FX F of X. So the derivative of F of X is going to be F of X. And F of X, in order to find this derivative of F of X, we must recall and use the product rule and chain rule. So when we do that, we will find that F of X is equal to D by DX of 2 X multiplied by the square root of 6 minus X plus 2 X multiplied by D by DX of the square root of 6 minus X. Fantastic. So when we reform the derivative, we will find that F of X is going to be equal to 2 multiplied by the square root of 6 minus X plus 2 X multiplied by -1 divided by 2 multiplied by the square root of 6 minus X. So I'm gonna skip a few steps, but essentially we need to take this expression for F of X, and we need to simplify it down to its most simple form or we can no longer simplify it, and using a little bit of algebra and moving things around, F of X will be ultimately equal to 12 minus 3 X divided by the square root of 6 minus X. Fantastic. So now our next step is we need to set F of X X equal to 0. So F of X equal to 0, so meaning we need to take 12 minus 3 x divided by the square root of 6 minus X is equal to 0. And now we need to rearrange this equation to isolate and solve for X all by itself. And when we do that, I'm going to skip a few steps again. So using a little bit of algebra to rearrange this equation once again to isolate and solve for X, you will find that X is equal to 12 divided by 3, which is equal to 4. Fantastic. So now our 4th step is we need to verify our critical point. So the critical point X equals 4 lies within our domain, and now we need to check if this is the only local extremium. So, once again, our critical point is X equals 4. And it lies within our domain of negative infinity, 6 square bracket. Fantastic. So now that we know that it lies within our interval, we now need to check if this is the only like the only value for the local extremum. So let us now determine the nature of the extremum using the first derivative test, or rather, we can check the behavior around the critical point. So, let us note that when X is less than 4, and if we choose a value of X equals 3, we will find that F 3 is equal to 12. Minus 3 multiplied by 3 divided by the square root of 6 minus 3. Once again, we're plugging in a value of 3 in for X, and when we do that, we will find that it's equal to 3 divided by the square root of 3. And 3 divided by the square root of 3 is indeed greater than 0. And for when X is greater than 4, let us say that X is equal to 5. So instead of rewriting our equation for like 12 minus 3 multiplied by X divided by the square root of 6 minus X, I'm just gonna give you the answer when we evaluate. And when we evaluate for X is greater than 4, and we choose X is equal to 5, we will find that F. Of 5 is going to be equal to -3 divided by the square root of 1, which is less than 0, and since F of X changes from a positive to a negative at X equals 4, that will mean that there's a local maximum at X equals 4. So now, our 5th and final step, which is our conclusionary steps, we need to note that according to the theorem, since there is exactly 1 local extremum, Extremum value, which is at X equals 4, and it's local maximum. F of 4 is the absolute maximum on the interval parenthesis negative infinity 6 square bracket. So when we calculate F of 4 is equal to 2 multiplied by 4 multiplied by the square root of 6 minus 4, we will find that this will be equal to when we evaluate it, 8 multiplied by the square root of 2. So, therefore, without a doubt, let's look at the look at our multiple choice answers. So therefore, without a doubt, The absolute maximum value is 8 multiplied by the square rooted 2 at X equals 4. So, therefore, our final answer, without a doubt, once again has to be multiple choice answer letter C, which is the absolute maximum of 8 multiplied by the square to 2 at X equals 4. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Verify that the following functions satisfy the conditions of Theorem 4.9 on their domains. Then find the location and value of the absolute extrema guaranteed by the theorem.

f(x) = x√(3-x)

Key Concepts

Theorem 4.9 (Extreme Value Theorem)

Continuity of Functions

Finding Extrema

Watch next