# Question Video: Finding the Absolute Maximum and Minimum Values of a Root Function in a Given Interval Mathematics • Higher Education

Find, if they exist, the values of the absolute maximum and/or minimum points for the function 𝑓(𝑥) = √(3𝑥 + 10) where 𝑥 ∈ [−2, 5].

06:11

### Video Transcript

Find, if they exist, the values of the absolute maximum and/or minimum points for the function 𝑓 of 𝑥 is equal to the square root of three 𝑥 plus 10, where 𝑥 is in the closed interval from negative two to five.

The question gives us a composite function, and it wants us to find the values of the absolute maximum and/or minimum if they exist, where 𝑥 is in the closed interval from negative two to five. We first want to check where our function 𝑓 of 𝑥 is continuous. We’re going to use the fact that it’s a composite function. We know if we have two continuous functions 𝑔 and ℎ, then their composition, 𝑔 composed with ℎ of 𝑥, is continuous on its domain. We see that our function 𝑓 of 𝑥 is the composition of two functions. It’s the composition of 𝑔 of 𝑥 is the square root of 𝑥 and ℎ of 𝑥 is equal to three 𝑥 plus 10.

And we know that both of these are continuous. The square root function is continuous, and all polynomials are continuous. So, because 𝑓 of 𝑥 is the composition of two continuous functions, it’s continuous on its domain. This just means 𝑓 of 𝑥 is continuous wherever 𝑓 of 𝑥 is defined. And we know the square root of a number will always be defined unless that number is negative. Therefore, our function 𝑓 of 𝑥 will not be defined when three 𝑥 plus 10 is less than zero, which also means it won’t be continuous for these values of 𝑥. We could subtract 10 from both sides of this inequality and then divide by three to see that our function 𝑓 of 𝑥 is not defined when 𝑥 is less than negative 10 over three.

However, we can see that this is the same as saying 𝑥 is less than negative three and one-third. And we see this is outside of the interval given to us in the question. So, we must have that our function 𝑓 of 𝑥 is continuous for all values of 𝑥 in the closed interval from negative two to five. As an aside, it’s worth noting at this point, since we’ve shown that our function 𝑓 of 𝑥 is continuous on a closed interval. By applying the extreme value theorem, we know that our function 𝑓 of 𝑥 will achieve a maximum and a minimum on the closed interval from negative two to five. So at this point, we already know that both of these values must exist.

Now, since our function 𝑓 of 𝑥 is continuous on the closed interval, we can find the values of the absolute maximum and absolute minimum by using the following three steps. First, we need to find the critical points of our function 𝑓 of 𝑥. That’s where the derivative is equal to zero or where the derivative does not exist. Next, we need to evaluate the function 𝑓 of 𝑥 at the critical points. Then, we need to evaluate the function 𝑓 of 𝑥 at the end points. The largest of these values will then be the absolute maximum of our function 𝑓 of 𝑥 on the interval, and the lowest will be the minimum of our function 𝑓 of 𝑥 on the interval.

So, the first thing we need to do is find the critical points of our function 𝑓 of 𝑥. We need to differentiate the square root of three 𝑥 plus 10. We know that our function 𝑓 of 𝑥 is a composite function. So, to differentiate it, we need to use the chain rule. We’ll set 𝑢 equal to the inner function of three 𝑥 plus 10. So, we have that 𝑓 of 𝑢 is equal to the square root of 𝑢, and 𝑢 of 𝑥 is three 𝑥 plus 10. 𝑓 is a function of 𝑢, and 𝑢 in turn is a function of 𝑥. Now, the chain rule tells us that if 𝑓 is a function of 𝑢 and 𝑢 in turn is a function of 𝑥, then the derivative of 𝑓 with respect to 𝑥 is equal to the derivative of 𝑓 with respect to 𝑢 multiplied by the derivative of 𝑢 with respect to 𝑥.

So, by applying the chain rule, we have the derivative of our function 𝑓 with respect to 𝑥 is equal to the derivative of the square root of 𝑢 with respect to 𝑢 multiplied by the derivative of three 𝑥 plus 10 with respect to 𝑥. We can differentiate the square root of 𝑢 with respect to 𝑢 by using the power rule for differentiation. We multiply by the exponent and then reduce the exponent by one. Since the square root of 𝑢 is the same as 𝑢 to the power of a half, this gives us one divided by two root 𝑢.

We can also differentiate three 𝑥 plus 10 by using the power rule for differentiation. It gives us three. So, the derivative of 𝑓 with respect to 𝑥 is equal to three divided by two root 𝑢. And we want this in terms of 𝑥, so we substitute 𝑢 is equal to three 𝑥 plus 10 to show that the derivative of 𝑓 with respect to 𝑥 is equal to three divided by two times the square root of three 𝑥 plus 10.

Now, we’re looking for the values of 𝑥 in our interval where the derivative is equal to zero or the derivative does not exist. We see that the positive square root of three 𝑥 plus 10, if it exists, will always give a number greater than or equal to zero. This is because we’re taking the positive square root of the number. If our denominator was equal to zero, then the derivative would not exist. However, if the square root of three 𝑥 plus 10 was positive, then multiplying it by two would give us a positive number. And then three divided by a positive number would give a positive number.

However, we’re only interested in 𝑥 is in the closed interval from negative two to five. So, in particular, 𝑥 must be greater than or equal to negative two which means that three 𝑥 plus 10 is never equal to zero. So, our derivative is the quotient of two positive numbers and that means it’s positive. So, on our interval, the derivative is never equal to zero. In fact, if our derivative is the quotient of two positive numbers, that means, in particular, it must exist on this interval. So, what we’ve shown is that our function has no critical points on the closed interval from negative two to five. This means we can just skip the second step because there are no critical points to evaluate our function at. All we need to do now is to evaluate our function 𝑓 of 𝑥 at the end points of our interval.

So, we have that 𝑓 evaluated at negative two is equal to the square root of three times negative two plus 10 which is the square root of four, which is just two. And 𝑓 evaluated at five is equal to the square root of three times five plus 10 which is the square root of 25 which is just equal to five. And since our function 𝑓 is continuous on this closed interval, it must have its minimum at two on this interval and its maximum at five on this interval. In conclusion, we have shown that the function 𝑓 of 𝑥 is equal to the square root of three 𝑥 plus 10, where 𝑥 is in the closed interval from negative two to five, will achieve an absolute minimum value of two and an absolute maximum value of five.

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