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

Find, if any, the local maximum and local minimum values of 𝑦 = 7π‘₯ + (7/π‘₯).

07:08

Video Transcript

Find, if any, of the local maximum and local minimum values of the function 𝑦 equal seven π‘₯ plus seven over π‘₯.

Firstly, we recall that local maxima and local minima are examples of critical points of a function. And the critical points of a function occur when its first derivative, in this case, d𝑦 by dπ‘₯, is either equal to zero or does not exist. It’s undefined. We, therefore, need to find an expression for the first derivative of this function. We will find it helpful, though, to rewrite our function 𝑦 slightly to begin with, or, more specifically, the second term. We can use laws of exponents to rewrite seven over π‘₯ as seven multiplied by π‘₯ to the power of negative one.

We can then see that our function 𝑦 is the sum of two general power terms. And so, we can recall the power rule of differentiation, which tells us that the derivative with respect to π‘₯ of π‘Žπ‘₯ to the 𝑛th power for real values of π‘Ž and 𝑛 is equal to π‘Žπ‘› multiplied by π‘₯ to the 𝑛 minus oneth power. We multiply by the original exponent 𝑛 and then reduce the exponent by one. Now, depending how much practice you’ve had of differentiation, you may find it helpful to think of seven π‘₯ as seven π‘₯ to the first power. And then, we can apply the power rule of differentiation.

Differentiating seven π‘₯ to the first power gives seven multiplied by one multiplied by π‘₯ to the zeroth power, which we’ll simplify in a moment. And then, differentiating seven π‘₯ to the power of negative one gives seven multiplied by negative one multiplied by π‘₯ to the power of negative two. Now, recall that π‘₯ to the power of zero is just one. So, our first term can be rewritten as seven. It’s just seven multiplied by one. And then, to the second term, we can rewrite this as negative seven over π‘₯ squared. So, we have our expression for the first derivative d𝑦 by dπ‘₯.

To find the π‘₯-values at any critical points, we then set this expression for d𝑦 by dπ‘₯ equal to zero, giving seven minus seven over π‘₯ squared equals zero. We can multiply through by π‘₯ squared in order to eliminate the denominator in the fraction, giving seven π‘₯ squared minus seven equals zero. We then divide through by seven and add one to each side of the equation to give π‘₯ squared is equal to one. Finally, we need to take the square root of each side of the equation, remembering that we need plus or minus the square root. So, we have π‘₯ is equal to plus or minus the square root of one, which is simply positive or negative one.

We’ve found then that there were two π‘₯-values at which our function has critical points. We also need to determine the value of the function itself for each of these π‘₯-values. Substituting into our original function 𝑦 then, which was seven π‘₯ plus seven over π‘₯, we have that when π‘₯ is equal to one, 𝑦 is equal to seven multiplied by one plus seven over one. That’s seven plus seven, which is equal to 14. And when π‘₯ is equal to negative one, we find that 𝑦 is equal to seven multiplied by negative one plus seven over negative one. That’s negative seven plus negative seven, which is equal to negative 14. We’ve found then that the two critical points of this function are at the point with coordinates negative one, negative 14 and the point with coordinates one, 14.

Finally, we need to determine whether these points are local minima or local maxima or perhaps points of inflection. And to do this, we’ll apply the second derivative test. To do this, we first find a general expression for the second derivative d two 𝑦 by dπ‘₯ squared, which we find by differentiating our first derivative again. Now, we may find it helpful to think of our first derivative as seven minus seven π‘₯ to the power of negative two. And then, we can apply the power rule of differentiation.

Again, depending how much practice you’ve had with differentiation, you may find it helpful to think of seven as seven π‘₯ to the power of zero. So when we apply the power rule of differentiation, we get seven multiplied by zero multiplied by π‘₯ to the negative one for the derivative of our first term. And then, we have negative seven multiplied by negative two multiplied by π‘₯ to the power of negative three. Of course, multiplying by zero just gives zero. So, this entire first term is zero. And we remember that the derivative of any constant is always zero. Our expression for the second derivative, therefore, simplifies to 14π‘₯ to the power of negative three, which we can write us 14 over π‘₯ cubed.

Finally, we need to evaluate the second derivative at each of our critical points. Substituting π‘₯ equals one, we find that at this point, d two 𝑦 by dπ‘₯ squared is equal to 14 over one cubed, which is equal to 14. Now, the key point here is that this is a positive value. And the second derivative test tells us that if the second derivative is positive at a critical point, then that critical point is a local minimum. So, we’ve found that the critical point one, 14 is a local minimum of our function 𝑦. At our other critical point, when π‘₯ is equal to negative one, we find that the second derivative of our function is equal to negative 14. So, this time, the second derivative is negative. And by the second derivative test, we find that this critical point is a local maximum.

So, by first finding the critical points of our function and then applying the second derivative test, we found that the local minimum value of the function 𝑦 is 14 and the local maximum value is negative 14. Remember that these local minimum and local maximum values are the values of the function itself at these critical points. Although coincidentally, they’re also the values of the second derivative at these critical points. But that is a coincidence. It’s the values of the function itself that we’re interested in.

Now, just one final point briefly, we said that critical points would occur when the first derivative is either zero or is undefined. And we haven’t mentioned anything about where this first derivative might be undefined. Well, if you recall, our first derivative was equal to seven minus seven over π‘₯ squared. And this will be undefined when the denominator of that fraction is zero, which will occur when π‘₯ is equal to zero. However, if we look at our original function 𝑦, we see that π‘₯ equals zero would also lead to the original function being undefined. And therefore, π‘₯ equals zero is not included in the domain of our function 𝑦. We, therefore, don’t need to be concerned that the first derivative does not exist at this point.

In summary then, we found that the local minimum value of our function is 14 and the local maximum value is negative 14.

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