In Exercises 1–10, find the extreme values (absolute and local...

Question

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

y = (𝓍 + 1) / (𝓍² + 2𝓍 + 2)

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says determine the extreme values absolute and local of the function G of X, which is equal to the quantity of 3 X plus 2 in quantity, divided by the quantity of X2 + 4 X + 5 in quantity over its natural domain. The first step in solving this problem is to determine the domain of our function GFX. And in order to determine our domain, we need to look at our function G of X and we noticed that we have a fraction. And our function G of X is not going to be defined whenever the denominator of a fraction is equal to 0, and the denominator is actually a quadratic function. So we need to determine whether this quadratic function is ever equal to 0, and we can do that by looking at the discriminate. And so the discriminant delta for this function is going to be equal to. 4 square minus 4 multiplied by 1 multiplied by 5. And when we evaluate this expression, this is going to be equal to -4. And since the discriminant here is negative, that means we don't have any real roots for this quadratic function. And so therefore, the denominator can never be zero in our function G of X, and therefore G of X is defined for all real numbers. So the domain of our function G of X is going to be the open interval from minus infinity to infinity. So now that we have the domain determined, we can now look at our critical points. And so, we're gonna need to calculate a derivative of our function G of X with respect to X. So we need to look for G of X in order to find a critical points. This is going to be equal to the derivative with respect to X. Of the quantity of 3 X plus 2 in quantity, divided by the quantity of X2 + 4 X plus 5 in quantity. So here I'm taking a derivative of a fraction, so we're going to want to use our quotient rule. So recall for the quotient rule, that if we have the function Y that has the form, Y is equal to U X divided by V of X. Then its derivative Y. is equal to the quantity of V multiplied by u minus U multiplied by V in quantity divided by V2d. So in our case, U is going to be equal to 3 X plus 2. And V is going to be equal to X2 + 4 X + 5. So U is going to be equal to the derivative with respect to X of the quantity of 3 X plus 2. And using our constant multiple and constant rule to take this derivative, this is going to be equal to 3. We also need to look at V, which is equal to the derivative with respect to X of the quantity of X2 + 4 X + 5 in quantity. And again, we're gonna need to use our constant multiple and constant rule, as well as our power rule to take this derivative. In doing so, we're going to get 2 X plus 4. So, using our quotient rule now, we'll see that G of X is going to be equal to the quantity of V of X, which is the quantity of X2 + 4 X. Plus 5 in quantity, multiplied by U, which is 3. We'll have minus U, which is the quantity of 3 X plus 2 in quantity, multiplied by V, which is the quantity of 2 X plus 4 in quantity, and all of that is divided by the quantity. Of X2. Plus 4 X + 5 in quantity squared. Now we just need to simplify this expression. We'll begin by multiplying everything out in the numerator. This is going to be equal to the quantity of 3 X2. Plus 12 acts. Plus 15 minus the quantity of 6 X. Plus 16 X. Plus 8 in quantity that is still all being divided by the quantity of X squared plus 4 X. Plus 5 in quantity squared. Now, distributing the negative sign and collecting like terms, this is going to be equal to. Minus 3 X2 minus. 4 X. Plus 7, and that whole quantity is divided by the quantity of X2 + 4 X. Plus 5 in quantity squared. And finally, factoring out a negative 1 out of the numerator, that's going to be equal to minus the quantity of 3 X2. Plus 4X Minus 7 in quantity, divided by the quantity of X2 + 4 X + 5 in quantity squared. So now that we have the first derivative calculated, we can use it to find our critical points. So recall that critical points are the X values at which our first derivative is either undefined or equal to zero. Now, in our case here, we again have a fraction, and we have the same denominator that we had in our original problem. So that means that The denominator here can never be equal to 0, so that means our function G of X is defined for all real numbers. So the only time that we can have critical points is when our first derivative here is equal to 0, and G of X is equal to 0 whenever the numerator is equal to 0. So we're going to set 3 x 2. Plus 4 X -7 equal to 0, and solve for X. Now, we can actually factor out the factor the left-hand side. And this factors to the quantity of 3 x plus 7 in quantity multiplied by the quantity of X minus 1 in quantity, and that has to be equal to 0. So we're going to set each factor equal to 0 and solve for X. We'll have 3 x plus 7 equal to 0. That means that X is equal to -7 divided by 3, and we'll have x minus 1 equal to 0, and so X is equal to 1. So now we need to determine whether these two critical points are either local minima, local maxima, or neither. So to do this, we're going to check to see if our function G of X is increasing or decreasing in the intervals just to the left and right of these critical points. So, the first interval that we're going to look at is going to be the interval going from minus infinity to -7 divided by 3. And so we're going to choose a point in this interval to calculate our first derivative at. So we're going to choose X equal to -4, so we'll calculate G of -4. And this is going to be equal to minus the quantity of 3 multiplied by -4 squared. Plus 4, multiplied by minus 4. -7 in quantity, divided by The quantity of -4 squad. Plus 4 multiplied by -4. Plus 5 in quantity where. And when we evaluate this expression, this is going to be equal to -1. And since our first derivative is negative over this interval, that means that our function is going to be decreasing over this interval. To recall that if the first root is negative, then our function is decreasing. So, the next interval that we want to look at is going to be from -7 divided by 3 to 1. And again, we'll just choose a point in this interval to calculate our first row, so we'll choose X equal to 0. We'll have G of 0, and this is going to be equal to minus the quantity of 3 multiplied by 0 squared, plus 4 multiplied by 0, minus 7 in quantity, divided by. The quantity of 0 squared. Plus 4 multiplied by 0, plus 5 in quantity squared, and when we evaluate this expression, this is going to be equal to 7 divided by 25. So here our um first derivative is positive. So since our first ro is positive, that means our function G of X is going to be increasing over this interval. And then the final interval that we want to look at is going to be the interval going from one to infinity. And again, we'll just choose a point to calculate our first root, so we'll choose X equal to 2. So we'll calculate G of 2, and that's going to be equal to minus. The quantity of 3 multiplied by 2 squared. Plus 4 multiplied by 2. Minus 7 in quantity, divided by the quantity of 2 squad. Plus or multiplied by 2. Plus 5 in quantity squared. And evaluating this expression, we get minus 13 divided by 200. 89. So, here, we see that our first derivative is negative, that means that our function is Decreasing over this interval. So now we can use this information to determine whether we have a local maximum or a local minimum. And so for X equal to minus 7 divided by 3. We see that to the left of that point, our function is decreasing, then after that point, it is increasing, so that means we have a local minimum at this point. And at X equal to 1, we see that our function is going from increasing to decreasing, which means that we have a local maximum. So now all we need to do is determine the values of our function G of X at these two points. So, we'll look at G minus 7 divided by 3. And that's going to be equal to the quantity of 3, multiplied by minus 7, divided by 3. Plus 2 in quantity, divided by the quantity of minus 7 divided by 3, that whole quantity squared. Plus or multiplied by minus 7 divided by 3. Plus 5 in quantity. And when we evaluate this expression, this is going to be equal to -4.5. And for our other point, We'll calculate G of 1. And that's going to be equal to the quantity of 3 multiplied by 1, plus 2 in quantity, divided by the quantity of 1 squared, plus 4 multiplied by 1. Plus 5, and when we evaluate this expression, this is going to be equal to 0.5. So, based on this information, We have a local. Minimum. At X equal to minus. 7 divided by 3. And that means that G of minus 7 divided by 3 is equal to -4.5. And we also have a local maximum. At X equal to 1 with G of 1. Equal to 0.5. So those are the local minimum and maximum that we were looking for for this problem. So now we just need to find the absolute maximum and absolute minimum of our function. And for this, we need to look at the limit as X purchase infinity. So if we look at the limit, As X approaches infinity. Of our function G of X, which is the quantity of 3 X plus 2 in quantity, divided by the quantity of X2 + 4 X plus 5 in quantity. And so, in order to take this limit, I'll first divi divide the numerator and denominator by X. So this is gonna be equal to the limit. As X approaches infinity. Of the quantity of 3 + 2 divided by X. In quantity, divided by the quantity of X plus 4 + 5 divided by X. And taking the limit as X purchases infinity, we're going to have the quantity of X divided by 2, and X 2 divided by X and 5 divided by X, those two quantities are going to go to 0 as X verses infinity. And so what I'm going to be left with is going to be equal to the limit. As X purchase infinity of 3 divided by X + 4. And as X approaches infinity here, our denominators can get really large, and so we'll have a constant divided by a really large number. So this is going to approach 0. So that means that this limit is going to be equal to 0. And likewise, if we look at the limit as X approaches minus infinity, Of the quantity of 3 X plus 2 divide in quantity, divided by the quantity of X2 + 4 X plus 5 in quantity, that this limit is also going to be equal to 0 by the same reasoning. So, since our function is approaching 0 for very large values of X, that means that our local minimum and local maximum are also our absolute minimum and absolute maximum. So we're gonna have an Absolute Minimum. At X equal to -7 divided by 3, with G of -7 divided by 3. Equal to -4.5. And we're gonna have an absolute maximum. At X equal to 1, with G of 1. Equal to 0.5. And so those are the absolute maximum and absolute minimum that we were looking for for this problem. So those are the answers that we were looking for. So, I hope that this explanation has been useful, and I'll see you in the next video.

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

y = (𝓍 + 1) / (𝓍² + 2𝓍 + 2)

Key Concepts

Critical Points

First Derivative Test

Domain of a Function

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