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

Find where (if at all) the function 𝑦 = βˆ’π‘₯ βˆ’ (1/(π‘₯ + 8)) has its local maxima and minima.

05:31

Video Transcript

Find where, if at all, the function 𝑦 equals negative π‘₯ minus one over π‘₯ add eight has its local maxima and minima.

Let’s first of all remind ourselves what local maxima and local minima are. The local maxima and minima are the largest and smallest values of the function within a given range. At these points, we can see that the slope, the derivative, is equal to zero. So we need to differentiate our function to find the slope function. And then we work out where the slope function is equal to zero. So we need to look at differentiating this function. We can differentiate this term by term. So let’s start with negative π‘₯. We can differentiate this, knowing that this is just negative π‘₯ to the power of one, and apply the power rule. This gives us that the derivative of negative π‘₯ is negative one.

Now we move on to the second term, one over π‘₯ add eight. We start by writing it as a negative exponent, π‘₯ add eight to the power of negative one. And we have a really useful tool for differentiating a function raised to a power. It’s the chain rule. This tells us that the derivative with respect to π‘₯ of a function 𝑓 of π‘₯ raised to the power of 𝑛 is 𝑛 multiplied by the function of π‘₯ raised to the power of 𝑛 minus one multiplied by 𝑓 prime of π‘₯. And so, the derivative with respect to π‘₯ of π‘₯ add eight to the power of negative one is 𝑛 the power, which is negative one, multiplied by our function, π‘₯ add eight, raised to the power of 𝑛 minus one. So negative one minus one, which is negative two, multiplied by 𝑓 dash of π‘₯, which is the derivative of π‘₯ add eight with respect to π‘₯.

So what’s the derivative of π‘₯ add eight? Well, differentiating term by term, π‘₯ differentiates to one. And eight is a constant. So it differentiates to zero. So the derivative of π‘₯ add eight is just one. So this is just negative π‘₯ add eight to the power of negative two. We can rewrite this as negative one over π‘₯ add eight squared. So d𝑦 by dπ‘₯ equals negative one minus, because it was a minus in the question, negative one over π‘₯ add eight squared. Because we’re subtracting a negative term here, we can write this as an add. Remember, we said that local maxima and minima occur where d𝑦 by dπ‘₯ equals zero. So we set our derivative equal to zero and solve for π‘₯.

We can start by adding one to both sides and then multiplying both sides by π‘₯ add eight squared. From here, we can square root both sides. And we must be careful here as the square root of one can take the value positive one or negative one. So when π‘₯ add eight equals positive one, π‘₯ equals negative seven. And when π‘₯ add eight equals negative one, π‘₯ equals negative nine. So now, we have the π‘₯-values of possible maxima or minima. But we don’t know which is maxima or minima. So let’s find out. I’m going to clear some space to do this.

Let’s start by finding the corresponding 𝑦-values by substituting our π‘₯-values into our original equation for 𝑦. So for our points negative seven, six and negative nine, 10, let’s check for maxima or minima. To check this, we evaluate the values of π‘₯. We found at the second derivative. If the second derivative is positive, π‘₯ is a minima. And if the second derivative is negative, π‘₯ is a maxima. To find the second derivative, we differentiate the first derivative. Negative one is a constant. So this differentiates to zero. For one over π‘₯ add eight squared, we can proceed in a similar way to how we found the first derivative. We rewrite this as a negative exponent and use the chain rule formula.

So the derivative of π‘₯ add eight to the power of negative two is negative two multiplied by π‘₯ add eight to the power of negative three multiplied by the derivative of π‘₯ add eight, which is one. And so π‘₯ add eight to the power of negative two differentiates to negative two multiplied by π‘₯ add eight to the power of negative three, which we can write as negative two over π‘₯ add eight cubed. And so the second derivative is negative two over π‘₯ add eight cubed.

I’m just going to clear some space here. So at π‘₯ equals negative seven, the second derivative is equal to negative two over negative seven add eight cubed, which is negative two over one cubed, which is just negative two. This is less than zero. So at π‘₯ equals negative seven, we have a maximum. At π‘₯ equals negative nine, the second derivative is equal to negative two over negative nine add eight cubed, which is equal to negative two over negative one cubed. And remember that negative one cubed is negative one. So this gives us two. This is greater than zero. So at π‘₯ equals negative nine, we have a minimum. So we found our local maximum to be negative seven, six and our local minimum to be negative nine, 10.

And hence the local maximum value is six. And the local minimum value is 10.

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