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Gl_ G.

Calculus 1 / AB

9 months ago

Hi there. So for this problem we are given the following function F of ads is equal to 13 and this plus four times adds minus adds a square In the interval between zero and 5. So we need to determine the absolute minimum value and the absolute maximum value of this function. So to find the absolute maximum or minimum values of a continuous function as this one on a close in turbo. The first thing that we need to do is to find the values of the function at the critical numbers of dysfunction. So what we are going to do to find this critical values is to take the derivative with respect to ads of this function. Then we will find that this is well the first term is zero because we know that the derivative of a constant is zero. The derivative the derivative, the derivative of the second term is one because that is ad. So we obtained just simply four and the derivative of the last term is -2 times ads. And we set this equal to zero because we want to find the critical point. Now we're solving here for the ads. We obtained this and solving for X we obtain that apps should be equal to do so this is a critical point. So what we are going to do now is to evaluate the function and with the critical value to so from there we obtain 13 plus four times two minus two to the square. So from this we obtain a value off 17. So now to know if this is a minimum value or a maximum value. What we need to do is to the largest of the values. Well for the second part is we need to find the values of the function at the endpoints of the interval. So the endpoints in this case are zero and five. So we need to evaluate this function at those two points. So at zero, if we evaluate this at zero we will have that The second term and the last term is going to be zero. So we obtained from this immediately that at the zero is 13 and we now substitute at the .5 so we obtain 13 plus four times 5 and this -5 to the square is in our coca cola. And From this we obtain a value of eight. So as you can see this value is less than the value that we obtained for the critical point. So We can conclude that F evaluated at two that we know is 17 is the maximum value maximum And f evaluated of fight that is equal to eight is the minimum. So the solution for this problem is that the maximum is 17 and the minimum is eight. So that the solution for this

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