Question Video: Determing the Value of the Absolute Maximum of a Rational Function on a Closed Interval Mathematics • Higher Education

True or False: The function 𝑓(π‘₯) = 2π‘₯/(1 + π‘₯Β²) has an absolute maximum over the closed interval [βˆ’3, 3] equal to 1.

06:20

Video Transcript

True or False: The function 𝑓 of π‘₯ is equal to two π‘₯ divided by one plus π‘₯ squared has an absolute maximum over the closed interval from negative three to three equal to one.

In this question, we’re given a function 𝑓 of π‘₯. And we need to determine if this function has an absolute maximum value over the closed interval from negative three to three equal to one. To answer this question, let’s start by determining if this function even has an absolute maximum on this interval. And to do this, we notice we’re given a closed interval. So we can try and do this by using the extreme value theorem.

We recall the extreme value theorem tells us that real-valued continuous functions on a closed interval will attain a maximum and a minimum value on this interval. And this is particularly useful because it helps us find the absolute maximum and minimum values on this interval. And we can apply this theorem in this case. First, we can see we have the closed interval from negative three to three. Next, we can see our function 𝑓 of π‘₯ is a rational function. And rational functions are continuous everywhere except when their denominator is equal to zero. And the denominator in our function 𝑓 of π‘₯ is never equal to zero because π‘₯ squared is greater than or equal to zero. So one plus π‘₯ squared is greater than or equal to one.

So 𝑓 of π‘₯ is continuous for all real values of π‘₯. And in particular, it’s continuous on the given closed interval. Therefore, the extreme value theorem tells us that 𝑓 of π‘₯ does indeed have an absolute maximum on this interval. But that’s not all we need to do. We also need to check that the output value of this maximum is equal to one. And to do this, we’ll clear some space and find the absolute maximum of this function over this interval.

First, we recall that the local extreme of 𝑓 of π‘₯ must occur at the critical points of 𝑓 of π‘₯. And the critical points are the points where the derivative of our function is equal to zero or where the derivative of our function does not exist. Next, we evaluate 𝑓 of π‘₯ at these points to find the values of any possible local extrema. And finally, we evaluate 𝑓 of π‘₯ at the endpoints of our interval because these can be possible absolute extrema.

So let’s start by finding the critical points of 𝑓 of π‘₯. To do this, we’re going to need to differentiate our function. And since 𝑓 of π‘₯ is a rational function, we’ll do this by using the quotient rule. We recall this tells us for two differentiable functions 𝑒 of π‘₯ and 𝑣 of π‘₯, the derivative of their quotient is equal to 𝑒 prime of π‘₯ times 𝑣 of π‘₯ minus 𝑣 prime of π‘₯ times 𝑒 of π‘₯ all divided by 𝑣 of π‘₯ all squared.

So to apply the quotient rule to 𝑓 of π‘₯, we set 𝑒 of π‘₯ to be the function in our numerator, that’s two π‘₯, and 𝑣 of π‘₯ to be the function in the denominator, that’s one plus π‘₯ squared. We then need to find the derivative of both of these functions. Let’s start with 𝑒 prime of π‘₯. That’s the derivative of the linear function two π‘₯. And the derivative of any linear function is the coefficient of π‘₯, which in this case is two.

Next, we need to find 𝑣 prime of π‘₯. We can do this term by term by using the power rule for differentiation. We need to multiply by the exponent of π‘₯ and reduce this exponent by one. Applying this, we get 𝑣 prime of π‘₯ is two π‘₯.

We’re now ready to find 𝑓 prime of π‘₯ by using the quotient rule. We need to substitute our expressions for 𝑒, 𝑣, 𝑒 prime, and 𝑣 prime into the quotient rule. We get that 𝑓 prime of π‘₯ is equal to two times one plus π‘₯ squared minus two π‘₯ multiplied by two π‘₯ all divided by one plus π‘₯ squared all squared. Distributing over the parentheses in our numerator, we get two plus two π‘₯ squared minus four π‘₯ squared all divided by one plus π‘₯ squared all squared. And finally, we can simplify our numerator. Two π‘₯ squared minus four π‘₯ squared is equal to negative two π‘₯ squared, giving us that 𝑓 prime of π‘₯ is equal to two minus two π‘₯ squared all divided by one plus π‘₯ squared all squared.

And now we’re ready to use our expression for 𝑓 prime of π‘₯ to find the critical points of 𝑓 of π‘₯. These are the points where 𝑓 prime of π‘₯ is equal to zero or 𝑓 prime of π‘₯ does not exist. Let’s start by checking if there’s any value of π‘₯ where 𝑓 prime of π‘₯ doesn’t exist.

We found the derivative of 𝑓 prime of π‘₯ by using the quotient rule. And the quotient rule is valid provided 𝑒 and 𝑣 are differentiable at that value of π‘₯ and 𝑣 of π‘₯ is not equal to zero. However, 𝑣 of π‘₯ was the denominator of our function 𝑓 of π‘₯. And we’ve already checked that this is never equal to zero for any value of π‘₯. So the quotient rule is valid for all real values of π‘₯. Therefore, 𝑓 prime of π‘₯ is defined for all real values of π‘₯.

So our only critical points will be when 𝑓 prime of π‘₯ is equal to zero. So let’s determine if there are any values of π‘₯ where 𝑓 prime of π‘₯ is equal to zero. For the quotient of two numbers to be equal to zero, the numerator must be equal to zero. And of course we can see the denominator is never equal to zero. Therefore, the critical points of 𝑓 of π‘₯ occur when this numerator is equal to zero.

We need to solve two minus two π‘₯ squared equals zero. And we can solve this. We add two π‘₯ squared to both sides of the equation, divide through by two, and then take the square root, where we remember we will get a positive and a negative root. We have π‘₯ is equal to positive or negative one. Therefore, we found all of the critical points of 𝑓 of π‘₯. There’s two: one when π‘₯ is equal to negative one and one when π‘₯ is equal to one.

So we can move on to the second part of our question. We need to evaluate 𝑓 of π‘₯ at its critical points. Let’s start with 𝑓 evaluated at one. We substitute one into our expression for 𝑓 of π‘₯ to get two times one divided by one plus one squared, which if we evaluate is equal to one. We can do the same to evaluate 𝑓 at its other critical point. We substitute π‘₯ is equal to negative one to get two times negative one all divided by one plus negative one all squared, which if we evaluate we get negative one.

Now that we’ve evaluated 𝑓 at all of its critical points, we need to move on to our third step. We need to evaluate 𝑓 at the endpoints of our interval. That’s negative three and three. Let’s start by evaluating 𝑓 at three. We substitute three into our expression to get two times three divided by one plus three squared, which is equal to 0.6. And we can do the same to evaluate 𝑓 at negative three. We see this is equal to negative 0.6.

And now that we’ve evaluated 𝑓 at all of its critical points and its endpoints, we know the largest of these values will be the absolute maximum of our function over the interval and the smallest will be the absolute minimum of the function over this interval. And the largest of these four values is one. So 𝑓 has an absolute maximum value of one on the closed interval from negative three to three, which is what we’re told in the question.

Therefore, we’ve shown that it’s true that the function 𝑓 of π‘₯ is equal to two π‘₯ divided by one plus π‘₯ squared has an absolute maximum value of one on the closed interval from negative three to three.

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