### Video Transcript

Find, if any, the local maximum and minimum values of the function π of π₯ is equal to five π₯ divided by 13 times π₯ squared plus one, together with their type.

The question gives us a function π of π₯ which is a rational function, and it wants us to find the local maximum and minimum values of this function π of π₯, together with the type of extrema that these values are. To start, we recall that local extrema will always occur at the critical points of our function. And we say that π₯ is equal to π is a critical point of the function π of π₯ if either the first derivative of π evaluated at π is equal to zero or the derivative does not exist at this value of π₯. So to find the critical points of our function π of π₯, weβre going to need to differentiate π of π₯. And our function π of π₯ is a rational function. Itβs the quotient of polynomials. So weβll differentiate this by using the quotient rule for differentiation.

We recall the quotient rule tells us the derivative of the quotient of two functions π’ over π£ is equal to π£ times π’ prime minus π’ times π£ prime all divided by π£ squared. So to apply the quotient rule to our function π of π₯, weβll set π’ of π₯ to be the polynomial in our numerator. Thatβs five π₯. And weβll set π£ of π₯ to be the polynomial in our denominator. Thatβs 13 times π₯ squared plus one. Using this, weβve rewritten π of π₯ as π’ of π₯ divided by π£ of π₯. So by the quotient rule, π prime of π₯ is equal to π£ of π₯ times π’ prime of π₯ minus π’ of π₯ times π£ prime of π₯ all divided by π£ of π₯ squared.

So to use the quotient rule, weβll need to find expressions for π’ prime of π₯ and π£ prime of π₯. Letβs start with π’ prime of π₯. Thatβs the derivative of five π₯ with respect to π₯. This is a linear function. So its derivative is the coefficient of π₯, which in this case is five.

Letβs now find an expression for π£ prime of π₯. Thatβs the derivative of 13 times π₯ squared plus one with respect to π₯. Weβll start by distributing 13 over our parentheses. This gives us 13π₯ squared plus 13. Now, we can just differentiate this by using the power rule for differentiation. We multiply by the exponent of π₯ and reduce this exponent by one. This gives us 26π₯. Substituting in our expressions for π’, π£, π’ prime, and π£ prime, we get that π prime of π₯ is equal to 13 times π₯ squared plus one times five minus five π₯ times 26π₯ all divided by 13 times π₯ squared plus one squared.

We now want to simplify this expression. Weβll start by distributing the square over the parentheses in our denominator. This gives us a new denominator of 13 squared times π₯ squared plus one squared. The next thing we want to do is cancel a shared factor of 13 in our numerator and our denominator. We now want to simplify our numerator. Weβll start by distributing five over our first set of parentheses. This gives us five π₯ squared plus five. Next, we have negative five π₯ multiplied by two π₯ is equal to negative 10π₯ squared. Finally, in our numerator, five π₯ squared minus 10π₯ squared is equal to negative five π₯ squared. So weβve shown π prime of π₯ is equal to five minus five π₯ squared divided by 13 times π₯ squared plus one squared. We found this expression of π prime of π₯ to find our critical points.

Remember, our critical points are where the derivative function is equal to zero or where the derivative function does not exist. Letβs start by checking the points where our derivative function does not exist. Since our derivative function is a rational function, it will exist for all real values of π₯ except where the denominator is equal to zero. So any solutions to the equation 13 times π₯ squared plus one squared is equal to zero will give us a critical point. However, we can show that there are no solutions to this equation. One way of doing this is to divide both sides of the equation by 13. Next, weβll take the square root of both sides of this equation. And since this is equal to zero, we donβt need to worry about adding a positive and negative sign.

Finally, weβll subtract one from both sides of the equation. So for our denominator to be equal to zero, we need π₯ squared to be equal to negative one, which has no real solutions. So our derivative function is defined for all real values of π₯. So the only critical points of our function will be where the derivative is equal to zero. So letβs now find all of the values of π₯ where our derivative function is equal to zero.

Since this is a rational function, it will only be equal to zero if the numerator is equal to zero and the denominator is not equal to zero. But we already showed our denominator is not equal to zero for any real value of π₯. So we just need to solve the equation five minus five π₯ squared is equal to zero. Weβll solve this equation for π₯. Weβll start by adding five π₯ squared to both sides of the equation. Next, weβll divide through by five. Finally, we can solve this by taking the square roots of both sides of this equation. We get that π₯ is equal to positive or negative one. So weβve shown our function π of π₯ has two critical points, one at π₯ is equal to one and the other at π₯ is equal to negative one.

We now need to see what types of points we have at these two values of π₯. We could do this by using the second derivative test. However, our first derivative is a rational function. So to do this, we would need to use the quotient rule a second time. So in this case, itβs easier to use the first derivative test. To use the first derivative test, we need to check the derivative of our function above and below each of our critical points. Our critical points are at negative one and one. So weβll check the derivative of π at negative two, zero, and two.

So letβs start filling in our table. We know negative one and one are critical points and the derivative of our function π is zero at these two points. Letβs now find π prime evaluated at negative two. Remember, weβre only interested in if this is a positive or a negative number. By evaluating this expression, we can see the denominator is positive and the numerator is negative. So this is a negative expression. So our derivative when π₯ is equal to negative two is negative. We can do the same to check our derivative when π₯ is equal to zero. Our denominator is positive and our numerator is positive. So our derivative is positive when π₯ is equal to zero.

Finally, we need to check the derivative of our function when π₯ is equal to two. If we evaluate these expressions, we can see our denominator is positive and our numerator is negative. So the derivative of our function π of π₯ is negative when π₯ is equal to two. So what does this mean for our critical points at negative one and one? For π₯ is equal to negative one, we have a negative derivative around π₯ is equal to negative two. At π₯ is equal to negative one, our derivative is zero. And our derivative goes positive around π₯ is equal to zero. So we can see by the first derivative test, at negative one, we have a local minima. We can do the same for π₯ is equal to one.

Around π₯ is equal to zero, our derivative is positive. At π₯ is equal to one, our derivative is equal to zero. And as π₯ approaches two, our derivative is negative. So we can see by the first derivative test, at π₯ is equal to one, we have a local maxima. Letβs now find the value of our local minima and local maxima. We have π evaluated at negative one is negative five divided by 26 and π evaluated at one is five divided by 26. We might be tempted to leave our answer like this. However, in this case, we can actually go further.

Letβs take a closer look at our derivative function. Weβve already explained the denominator of our derivative function is positive for all values of π₯. So letβs focus on the numerator, five minus five π₯ squared. Since this represents the slope of our function, when this slope is negative, our function is decreasing and when this slope is positive, then our function is increasing. So letβs think about what this means. If our value of π₯ is greater than one, then five minus five π₯ squared is less than zero. Our slope is negative. So our function is decreasing for all values of π₯ greater than one.

But if we look at our function π of π₯, we can see when π₯ is greater than one, our function is positive. So our function decreases for π₯ greater than one but stays positive. We can make a similar argument for π₯ is less than negative one. Our slope will be negative. And if our slope is negative when π₯ is less than negative one, our function is decreasing, which means for all values of π₯ less than negative one, our function π of π₯ is bigger than negative five over 26. And if we look at our function π of π₯, we can see when π₯ is less than negative one, our function is always negative. So for π₯ is less than negative one, our function is between zero and negative five over 26.

And we can say something similar for π₯ greater than one. π of π₯ is between five over 26 and zero. And weβve also shown that thereβs no critical points between π₯ is equal to negative one and π₯ is equal to one. So in actual fact, these are not just local extrema, they must be global extrema. Therefore, weβve shown that the function π of π₯ is equal to five π₯ divided by 13 times π₯ squared plus one has an absolute maximum of five over 26 and an absolute minimum of negative five divided by 26.