Video: Finding the Absolute Maximum and Minimum Values of a Rational Function on a Given Interval

Determine the absolute maximum and minimum values of the function 𝑦 = π‘₯/(2π‘₯ + 8) on the interval [2, 6].

03:01

Video Transcript

Determine the absolute maximum and minimum values of the function 𝑦 equals π‘₯ over two π‘₯ plus eight on the closed interval two to six.

Remember, to find absolute extrema for our function 𝑓 of π‘₯ over some closed interval, we follow three steps. We begin by finding any critical points in our closed interval. We then find the values of 𝑓 of π‘₯ at these critical points. And then, we check the end points for absolute extrema, in other words, values that are smaller than the relative minimum or larger than the relative maximum.

Remember, the critical points are the points on our curve where the derivative is equal to zero or possibly does not exist. So, we need to find the derivative of our function and set it equal to zero. But how do we differentiate π‘₯ over two π‘₯ plus eight? In fact, we have a number of methods we could use. But since it’s the quotient of two differentiable functions, we can use the quotient rule.

This says that the derivative of the quotient of two differentiable functions 𝑒 and 𝑣 is 𝑣 times d𝑒 by dπ‘₯ minus 𝑒 times d𝑣 by dπ‘₯ all over 𝑣 squared. The numerator of our fraction is π‘₯, so we’re going to let 𝑒 be equal to π‘₯ and 𝑣 be equal to two π‘₯ plus eight. Then, d𝑒 by dπ‘₯ is equal to one. And d𝑣 by dπ‘₯ is equal to two.

So, the derivative of 𝑦 with respect to π‘₯ is two π‘₯ plus eight times d𝑒 by dπ‘₯, which is one, minus 𝑒 times d𝑣 by dπ‘₯, that’s π‘₯ times two, all over 𝑣 squared, that’s two π‘₯ plus eight all squared. So, we can say that d𝑦 by dπ‘₯ is equal to eight over two π‘₯ plus eight all squared. Next, we’re going to set this equal to zero and solve for π‘₯. But look at what happens when we do.

For an algebraic fraction to be equal to zero, the numerator must itself be equal to zero. In this case, we end up with the statement zero equals eight, which we know to be rubbish. This means d𝑦 by dπ‘₯, in this case, cannot be equal to zero. There are no turning points for us to evaluate. And we, therefore, jump straight to step three and evaluate the function at the end points of the interval. That’s 𝑓 of two and 𝑓 of six.

𝑓 of two is two over two times two plus eight, that’s two twelfths, which simplifies to one-sixth. 𝑓 of six is six over two times six plus eight. That’s six twentieths, which simplifies to three-tenths. Three-tenths is greater than one-sixth, so we can say the absolute maximum value of our function is three-tenths and the absolute minimum value is one-sixth. Now let’s backtrack a little though.

We said that critical points occur at places on the function where the derivative does not exist, and there is one point on our function where the derivative doesn’t exist. That’s the point where two π‘₯ plus eight is equal to zero, or π‘₯ is equal to negative four. Now since this was outside the closed interval two to six, we actually didn’t need to worry about this critical point. And we could focus solely on the end points of the intervals 𝑓 of two and 𝑓 of six.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.