Question Video: Finding the Local Maximum and Minimum Values of a Function Mathematics • Third Year of Secondary School

Video Transcript

Determine, if there are any, the values of the local maximum and the local minimum of the function 𝑦 is equal to negative 𝑥 minus one over 𝑥 plus eight.

Let’s begin by recalling what we mean by a local maximum and a local minimum. The local maximum and minimum are the largest and smallest values within a given range. At these points, we see that the slope, or derivative, is equal to zero. This means that the first step in this question is to differentiate our function to find the slope function. We will then be able to find the point at which this is equal to zero. We will differentiate our function term by term. However, before we do this, it is worth recalling how we can rewrite the second term. One of our laws of exponents states that one over 𝑥 is equal to 𝑥 to the power of negative one. This means that our function can be rewritten as 𝑦 is equal to negative 𝑥 minus 𝑥 plus eight to the power of negative one.

We are now in a position to differentiate term by term, and we can differentiate negative 𝑥 using the power rule of differentiation, which states that differentiating 𝑥 to the power of 𝑛 with respect to 𝑥 is equal to 𝑛 multiplied by 𝑥 to the power of 𝑛 minus one. This means that differentiating negative 𝑥 to the power of one gives us negative one multiplied by 𝑥 to the power of zero. And since 𝑥 to the power of zero is equal to one, this simplifies to negative one.

We can differentiate the second term using the chain rule. This states that the derivative with respect to 𝑥 of a function 𝑓 of 𝑥 raised to the power of 𝑛 is equal to 𝑛 multiplied by 𝑓 of 𝑥 to the power of 𝑛 minus one multiplied by 𝑓 prime of 𝑥, where 𝑓 prime of 𝑥 is the derivative of 𝑓 of 𝑥. Differentiating 𝑥 plus eight to the power of negative one gives us negative one multiplied by 𝑥 plus eight to the power of negative two multiplied by one. And d𝑦 by d𝑥 is therefore equal to negative one minus this expression. Our expression simplifies to negative one plus 𝑥 plus eight to the power of negative two. And this can be rewritten as negative one plus one over 𝑥 plus eight squared.

Recalling that local maximum and local minimum points occur when d𝑦 by d𝑥 is equal to zero, we can set this expression equal to zero. To solve for 𝑥, we begin by adding one to both sides. We can then multiply through by 𝑥 plus eight all squared, giving us 𝑥 plus eight all squared is equal to one. Next, we square root both sides such that 𝑥 plus eight is equal to positive or negative the square root of one. This gives us two possible values: either 𝑥 plus eight is equal to one or 𝑥 plus eight is equal to negative one. Solving both of these equations by subtracting eight from both sides, we have 𝑥 is equal to negative seven or 𝑥 is equal to negative nine.

We now have 𝑥-values of possible local maximum and local minimum points. But how do we work out which is which? Well, let’s firstly work out the corresponding 𝑦-values. When 𝑥 is equal to negative seven, 𝑦 is equal to negative negative seven minus one over negative seven plus eight. This simplifies to seven minus one, which is equal to six. Repeating this process when 𝑥 is equal to negative nine, we have 𝑦 is equal to negative negative nine minus one over negative nine plus eight. This simplifies to nine plus one, which is equal to 10. When 𝑥 is equal to negative seven, 𝑦 is equal to six. And when 𝑥 is equal to negative nine, 𝑦 is equal to 10.

This means that we have two points that could be local maximum or local minimum points of the function. They have coordinates negative seven, six and negative nine, 10. The values of six and 10 could therefore correspond to local maximum or local minimum values. In order to identify which one they are, if either, we consider the second derivative. If d two 𝑦 over d𝑥 squared is positive, we have a local minimum, whereas if d two 𝑦 over d𝑥 squared is negative, we have a local maximum. If the second derivative is equal to zero, we have a possible point of inflection.

To find the second derivative, we will differentiate our expression d𝑦 by d𝑥 with respect to 𝑥. Once again, we can do this term by term, and we recall that differentiating a constant gives us zero. Using the chain rule to differentiate 𝑥 plus eight to the power of negative two with respect to 𝑥 gives us negative two multiplied by 𝑥 plus eight to the power of negative three multiplied by one, which simplifies to negative two over 𝑥 plus eight cubed. We can now substitute 𝑥 equals negative seven and 𝑥 equals negative nine into this expression. When 𝑥 is equal to negative seven, the second derivative is equal to negative two over negative seven plus eight cubed. This is equal to negative two. And as negative two is less than zero, we have a maximum point at negative seven, six.

Repeating this process when 𝑥 is equal to negative nine, the second derivative is equal to negative two over negative nine plus eight cubed. And this is equal to two. This time, as the second derivative is greater than zero, we have a local minimum point with coordinates negative nine, 10. We can therefore conclude that for the function 𝑦 is equal to negative 𝑥 minus one over 𝑥 plus eight, there is one local maximum value equal to six and one local minimum value equal to 10.