Question Video: Finding the Local Maximum and Minimum Values of a Function Mathematics

Video Transcript

Find the points 𝑥, 𝑦 where 𝑦 equals nine 𝑥 plus nine over 𝑥 has a local maximum or a local minimum.

Local maxima and local minima are examples of critical points. And we recall that at the critical points of a function, the first derivative d𝑦 by d𝑥 is equal to zero. Before differentiating, we may find it helpful to rewrite the second term in our function as nine 𝑥 to the negative one. We can then use the power rule of differentiation to find the first derivative d𝑦 by d𝑥. Remember that when we differentiate, we decrease the power by one. So when we decrease that power of negative one, it will become negative two not zero. Watch out for that! That’s a common mistake. We can rewrite this derivative as nine minus nine over 𝑥 squared and we’ll then set this derivative equal to zero.

We’ll now solve the resulting equation in order to find the 𝑥-values at the critical points. We begin by multiplying every term in the equation by 𝑥 squared. We can then divide through by nine to give 𝑥 squared minus one equals zero. Add one to both sides and then finally take the square root, remembering that we have both positive and negative solutions. We find that 𝑥 is equal to positive or negative one. So this function has two critical points.

Next, we need to find the 𝑦-values at each critical point by evaluating the function itself. When 𝑥 is equal to positive one, 𝑦 is equal to nine multiplied by one plus nine over one, which is equal to 18, giving a critical point of one, 18. When 𝑥 is equal to negative one, 𝑦 is equal to negative 18. So our second critical point has coordinates negative one, negative 18. We now need to determine whether these critical points are local minima or local maxima, which we’ll do using the second derivative test. We’ll clear some space in order to do this.

To find the second derivative d two 𝑦 by d𝑥 squared, we need to differentiate the first derivative, which was nine minus nine 𝑥 to the power of negative two with respect to 𝑥. Doing so, we obtain negative nine multiplied by negative two 𝑥 to the power of negative three, which we can write as 18 over 𝑥 cubed. Next, we need to evaluate this second derivative at each of our critical points. When 𝑥 is equal to negative one, the second derivative is 18 over negative one cubed, which is equal to negative 18. This is less than zero. And we recall that if the second derivative of a function is negative at a critical point, then the critical point is a local maximum. Evaluating the second derivative when 𝑥 is equal to positive one gives 18 over one cubed, which is 18. And as this is greater than zero, we conclude that the critical point when 𝑥 is equal to one is a local minimum.

So we’ve completed the problem. We answered that the point one, 18 is a local minimum and the point negative one, negative 18 is a local maximum.