Video: Finding the Coordinates of Local Extrema for a Rational Function

Find the coordinates of all the local minima and maxima of the function 𝑓(π‘₯) = (3/π‘₯) + 5 + 6π‘₯.

06:58

Video Transcript

Find the coordinates of all the local minima and maxima of the function 𝑓 of π‘₯ is equal to three divided by π‘₯ plus five plus six π‘₯.

The question is asking us to find all the coordinates of the local minima and maxima of the function 𝑓 of π‘₯. The first thing we need to remember is local extrema will always occur at critical points of our function. And remember, we say that there’s a critical point when π‘₯ is equal to π‘Ž if 𝑓 prime of π‘Ž is equal to zero or the derivative of 𝑓 does not exist when π‘₯ is equal to π‘Ž. The first thing we’ll do is check the domain of our function 𝑓 of π‘₯. The reason we do this is if 𝑓 of π‘₯ is not defined at a certain value of π‘Ž, then it certainly can’t have a local maxima or a local minima at this point.

We can see that the only part of our function that won’t be defined for some value of π‘₯ is when we divide by π‘₯. So our function 𝑓 of π‘₯ is not defined when π‘₯ is equal to zero. We know it’s defined for all other real values of π‘₯. So we don’t need to worry about the case when π‘₯ is equal to zero. Let’s now find all the critical points of our function 𝑓 of π‘₯. To find the critical points of our function 𝑓 of π‘₯, we’re going to want to differentiate 𝑓 of π‘₯. To do this, we rewrite this as three π‘₯ to the power of negative one plus five plus six π‘₯ to the first power.

We can now differentiate this term by term by using the power rule for differentiation. We want to multiply by our exponent of π‘₯ and reduce this exponent by one. This gives us negative three π‘₯ to the power of negative two plus six. And we reorder our terms and use our laws of exponents to write this as six minus three divided by π‘₯ squared. Now remember, we found this expression for 𝑓 prime of π‘₯ to find the critical points of our function 𝑓 of π‘₯. So let’s start by checking all values of π‘₯ where our derivative does not exist.

We can see the only values of π‘₯ where our derivative does not exist is when we’re dividing by zero. But remember, π‘₯ is equal to zero is not in the domain of our function 𝑓 of π‘₯. What this means is for all values of π‘₯ in the domain of our function 𝑓 of π‘₯, the derivative of our function exists. So the only possible critical points will occur when the derivative is equal to zero. So let’s now check for values of π‘₯ where our derivative is equal to zero. This means we want to find the values of π‘₯ where six minus three over π‘₯ squared is equal to zero.

We’ll add three over π‘₯ squared to both sides of this equation then multiply through by π‘₯ squared. Then, we’ll divide both sides of the equation by six. This gives us π‘₯ squared is equal to one-half. And then we can solve this by taking the square root of both sides of this equation. Remember, we’ll get a positive and a negative square root. This gives us π‘₯ is equal to positive or negative one divided by root two. And we can simplify this slightly by rationalizing our denominator. We’ll multiply both the numerator and denominator by root two. This gives us that one over root two is equal to root two divided by two. So, we found two critical points of our function 𝑓 of π‘₯, one when π‘₯ is equal to root two over two and the other when π‘₯ equal to negative root two over two.

But remember, the question wants us to find the coordinates of all the local maxima and local minima of this function. So we need to check whether these critical points are local extrema and then find the coordinates of all the local extrema. To check whether these critical points are local extrema, we’ll use the first derivative test. We’ll do this because we’ve already found an expression for 𝑓 prime of π‘₯. To use the first derivative test. We want to find the slope of our function to the left and to the right of our critical points. But we have to be careful in this case. Remember, our derivative function is not defined when π‘₯ is equal to zero. This means when we’re choosing our values of π‘₯ for each of our critical points, we don’t want to pick points either side of π‘₯ is equal to zero.

So to use the first derivative test to check π‘₯ is equal to negative root two over two, we chose two negative values of π‘₯. Similarly, when we were choosing our values of π‘₯ for π‘₯ is equal to root two over two in the first derivative test, we chose two positive values of π‘₯. So let’s start filling in our table. First, when π‘₯ is equal to negative root two over two and when π‘₯ is equal to root two over two, we have the slope of our function is equal to zero. Next, let’s find 𝑓 prime of negative one. We substitute π‘₯ is equal to negative one into our expression for 𝑓 prime of π‘₯. We get six minus three divided by negative one squared. And we can then evaluate this expression. We get three. So the slope of our function when π‘₯ is equal to negative one is positive.

Next, let’s find the slope of our function when π‘₯ is equal to negative one-half. We’ll substitute π‘₯ is equal to negative 0.5 into our expression for 𝑓 prime of π‘₯. We get six minus three divided by negative 0.5 squared. And if we evaluate this expression, we get negative six. So we’ve shown the slope of our function when π‘₯ is equal to negative one-half is negative. We can do something similar to find the slope of our function when π‘₯ is equal to one-half and the slope of our function when π‘₯ is equal to one. We get 𝑓 prime of 0.5 is negative six. So the slope of our function when π‘₯ is equal to one-half is negative, and 𝑓 prime of one is equal to three. So the slope of our function when π‘₯ is equal to one is positive.

So what does this table mean for our critical points? We see when π‘₯ is equal to negative one, our function is sloping upwards. However, when π‘₯ is equal to negative root two over two, our slope is equal to zero. And when π‘₯ is equal to negative one-half, our slope is negative. So we can see from this sketch, when π‘₯ is equal to negative root two over two, we have a local maximum. We can do the same for our other critical point. When π‘₯ is equal to negative one-half, our slope is negative. However, when π‘₯ is equal to root two over two, our slope is equal to zero. Finally, when π‘₯ is one, our slope is positive. We can then see from this sketch, when π‘₯ is equal to root two over two, we’ll have a local minimum.

So we’ve now shown both of our critical points are local extrema. All we need to do now is find the coordinates of our critical points. Let’s start with π‘₯ is equal to root two over two. We’ll substitute this into our expression for 𝑓 of π‘₯. We get three divided by root two over two plus five plus six times root two over two. Remember, root two over two is actually one divided by root two. So dividing three by one over root two is the same as multiplying three by one over root two. This means we can simplify our expression to three root two plus five plus three root two, which is, of course, equal to five plus six root two.

We’ll do the same to find the coordinates when π‘₯ is equal to negative root two over two. We substitute this into our expression for 𝑓 of π‘₯. We get three divided by negative root two over two plus five plus six times negative root two over two. We then simplify this in the same way we did before. We get negative three root two plus five minus three root two, which we can calculate; it’s equal to five minus six root two.

Therefore, we were able to show the function 𝑓 of π‘₯ is equal to three over π‘₯ plus five plus six π‘₯ has one local minima with coordinates root two over two, five plus six root two and one local maxima with coordinates negative root two over two, five minus six root two.

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