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) = 4x + 1/√x

Accepted Answer

Hello. In this video, we are going to be using the following theorem. The problem asks us to check if the conditions, or if the function F of X is equal to X + 2 divided by the square root of X. We want to check to see if this function 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. So the theorem says, suppose F is continuous on an interval I that contains exactly one local extrema at C. If a local maximum occurs at C, then F of C is the absolute maximum of F on the interval. If a local minimum occurs at C, then F of C is the absolute minimum of the of F on the interval. So, how exactly do we approach this problem? Well, let's go ahead and break down what the theorem is saying. So, the first part of the theorem is saying that the function F must be continuous on the given interval or on a interval. So, the first part that we need to do is we need to check the domain of the function. So, the function given to us is F of X equal to X + 2 divided by the square root of X. Now we know that the domain of X is all real numbers, but notice that we have a square root of X in the denominator. The domain of the square root of X in a denominator must be X strictly greater than 0. So that is the domain, but now we need to check to see if the function is continuous on the domain. Well, since 2 divided by square root of X has a domain of X greater than 0, and X is all real numbers, that means that this function will be continuous on its domain. Now what we want to do is you want to go ahead and find the critical points, because the second part of the theorem saying that we need to find the local maximums or the local minimums. So, the third part of this theorem requires us to find the local extrema. In order to do this, we will need to take the derivative of F of X. F of X is equal to 1. Plus 2 multiplied by the derivative of 1 divided by the square root of X, and that is going to be negative 1/2 multiplied by X, raises the power of -3 divided by 2. To further simplify this function, we can simplify this by multiplying the coefficients together, and F of X is equal to 1 minus X, raise the power of -3 halves. To find the local extrema, we want to find the values of X that cause the the derivative to be zero. In order to do this, we will take our derivative, 1 minus X, raise the power of -3 halves, set it equal to 0, and solve. Since we don't want to work with a negative exponent, the first thing we can do is multiply through the entire function or the entire equation by X rates the power of 3 halves. That will allow us to rewrite the equation to be X, raises the power of 3 halves, minus 1 equal to 0. If we add one to both sides of the equation, we get X rates to the power of three halves. Equal to 1 And this is going to further simplify to be X is equal to 1. What this means is that we have exactly 1 critical point. But now what we want to do is we want to find the local extrema using this critical point on the domain. Now, again, the domain is that X must be strictly greater than 0. So what we are going to do is we are going to create a number line. This number line will start at 0 but not include the value and go off to positive infinity. Then we will go ahead and mark our critical point, which is one on the number line. Now we will go ahead and use the first derivative test to determine the behaviors around the critical point of 1. We will go ahead and take the testing point 0.5, and plug this into the derivative. When we plug in 0.5 into the derivative, the derivative is approximately equal to -1.8, which tells us that the region between 0 and 1 will be decreasing for the graph of F of X. We will take a testing 0.2 on the interval from 1 to infinity. F of 2 will give us an approximate value of 0.6. Since this is a positive value, we know that the graph will be increasing on the interval from one to infinity. Since the behavior of the graph switches from decreasing to increasing by the first derivative test, this tells us that one. Or X equal to 1 is a local minimum. Now, let's go ahead and go back to the theorem. The theorem says that if a local maximum occurs at C, then F of C is the absolute maximum on F on the interval. And if a local minimum occurs at C, then F of C is the absolute minimum of F on the interval. Since we just identified this value, X equal to 1 to be a local minimum, what we want to do now is you want to go ahead and plug this into the original function to solve for the value of C of the local or the absolute minimum. So, plugging in 1 into the original function F 1 was equal to 1 + 2 divided by the square root of 1. This will further simplify to be 1 + 2, which is equal to 3, and by the theorem, the absolute maximum will occur at X's equal to 1 and have a value of 3. And the theorem will apply. And with that being said, the solution to this problem is going to be a So I hope this video helps you in understanding on how to use this theorem, and I will go ahead and see you all in the next video.

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) = 4x + 1/√x

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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) = 4x + 1/√x