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

Find the local maxima and local minima of 𝑓(π‘₯) = βˆ’(5π‘₯Β²/3) + 2π‘₯ βˆ’ (1/6) ln π‘₯, if any.

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Video Transcript

Find the local maxima and local minima of 𝑓 of π‘₯ is equal to negative five π‘₯ squared over three plus two π‘₯ minus one-sixth times the natural logarithm of π‘₯, if any.

We’re given a function 𝑓 of π‘₯. We need to determine all of the local maxima and local minima of this function if there are any. To find these points, we need to notice something about our function 𝑓 of π‘₯. We know how to differentiate 𝑓 of π‘₯. This is because 𝑓 of π‘₯ is the sum of a polynomial and the natural logarithm function, and we know how to differentiate both of these. We also know that local extrema of a function will always occur when the derivative of our function is equal to zero or where the derivative does not exist.

We also sometimes call these critical points. So we need to find an expression for 𝑓 prime of π‘₯ to find all of our critical points. We know 𝑓 prime of π‘₯ will be equal to the derivative of negative five π‘₯ squared over three plus two π‘₯ minus one-sixth times the natural logarithm of π‘₯ with respect to π‘₯. And we in fact know how to differentiate this term by term. The first two terms in this expression can be differentiated by using the power rule for differentiation. And to differentiate our third and final term, we need to remember the derivative of the natural logarithm with respect to π‘₯ is equal to the reciprocal function, one over π‘₯.

So we can now evaluate this derivative. Again, we differentiate the first two terms by using the power rule for differentiation. We multiply by our exponent of π‘₯ and reduce this exponent by one. We get that 𝑓 prime of π‘₯ is equal to negative 10π‘₯ divided by three plus two minus one divided by six π‘₯. We want to find all the critical points of our function 𝑓 of π‘₯. Remember, that’s where the derivative of 𝑓 of π‘₯ is equal to zero or where the derivative does not exist.

And we can see that 𝑓 prime of π‘₯ is the sum of rational functions. And the only time a rational function will not be defined is when the denominator is equal to zero. And the only denominator here which can equal zero is six π‘₯. So 𝑓 prime of π‘₯ will exist for all values of π‘₯ except when six π‘₯ is equal to zero, which is of course when π‘₯ is equal to zero.

But before we carry on, there’s one more thing we need to do. We need to check that π‘₯ is equal to zero is in the domain of our function 𝑓 of π‘₯. This is because we can’t possibly check the derivative of our function at zero if our original function is not even defined when π‘₯ is equal to zero. And in fact we can see that π‘₯ is equal to zero is not in the domain of 𝑓 of π‘₯. Otherwise, we would need to take the natural logarithm of zero. So zero is not a critical point of our function 𝑓 of π‘₯. And in fact, this tells that 𝑓 prime of π‘₯ is defined for all values of π‘₯ on the domain of 𝑓 of π‘₯.

So the only critical points of our function will be where the derivative of 𝑓 of π‘₯ is equal to zero. One way of solving this equation is to multiply the entire equation through by π‘₯. Remember, we’ve already explained that π‘₯ cannot be equal to zero. Multiplying through by π‘₯ and simplifying, we get that 𝑓 prime of π‘₯ will be equal to zero when negative 10π‘₯ squared over three plus two π‘₯ minus one-sixth is equal to zero. And this is just a quadratic in π‘₯, so we could solve this by using a quadratic solver or the quadratic formula.

Another way of solving this is to multiply our equation through by negative six. We get 20π‘₯ squared minus 12π‘₯ plus one is equal to zero. And then, either by inspection or the factor theorem or one of the methods we mentioned previously, we can factor this to give us 10π‘₯ minus one times two π‘₯ minus one is equal to zero. And we know if the product of two factors is equal to zero, then one of these two factors must be equal to zero. In other words, either 10π‘₯ minus one is equal to zero or two π‘₯ minus one is equal to zero.

And then we can just solve both of these linear equations for π‘₯. We get either π‘₯ is equal to one over 10 or π‘₯ is equal to one-half. And just as we did before, we should also check that our function 𝑓 of π‘₯ is defined when π‘₯ is equal to one over 10 and when π‘₯ is equal to one-half. And if we did this, we can see, of course, we can substitute π‘₯ is equal to one over 10 or π‘₯ is equal to one-half into the polynomial part of 𝑓 of π‘₯. And both of these two values are positive, so we can also substitute them into the natural logarithm of π‘₯. So both of these values are in the domain of 𝑓 of π‘₯.

Therefore, both of these points are critical points of our function 𝑓 of π‘₯. The derivative of 𝑓 of π‘₯ at these points is equal to zero, which of course tells us that these will be local extrema of our function. But we still need to determine whether these are local maxima or local minima and we need to find their values. And there’s a few different ways we could do this. For example, we could use the first derivative test. However, if we look at our expression for the first derivative of 𝑓 of π‘₯ with respect to π‘₯, we see we can differentiate this one more time. This means we could also use the second derivative test.

To use the second derivative test, we’re going to need to find an expression for the second derivative of 𝑓 with respect to π‘₯. We can do this by differentiating 𝑓 prime of π‘₯ with respect to π‘₯. That’s the derivative of negative 10π‘₯ over three plus two minus one over six π‘₯ with respect to π‘₯. And to make this easier to differentiate, we’ll rewrite negative one over six π‘₯ as negative one-sixth multiplied by π‘₯ to the power of negative one. This means we can now evaluate this derivative term by term by using the power rule for differentiation. And then by applying the power rule of differentiation term by term, we get negative 10 over three plus one-sixth times π‘₯ to the power of negative two.

And of course by using our laws of exponents, we can rewrite π‘₯ to the power of negative two in our denominator as π‘₯ squared. This gives us that 𝑓 double prime of π‘₯ is equal to negative 10 over three plus one over six π‘₯ squared. We now need to recall the second derivative test for a function 𝑓 of π‘₯. We recall this tells us if π‘₯ is a critical point of the function 𝑓 and the second derivative of 𝑓 with respect to π‘₯ at π‘₯ is positive, then it must be a local minima. However, if the second derivative of 𝑓 with respect to π‘₯ at π‘₯ is negative, then it must be a local maxima.

So we now need to evaluate the second derivative of 𝑓 of π‘₯ with respect to π‘₯ at each of our critical points. We’ll start with π‘₯ is equal to one over 10. We’ll write this as 0.1. Substituting 0.1 into our expression for 𝑓 double prime of π‘₯, we get negative 10 over three plus one divided by six times 0.1 squared. And if we evaluate this expression, we get 40 divided by three, and we need to notice this is positive. And because the second derivative of 𝑓 of π‘₯ with respect to π‘₯ at 0.1 is positive, this means around this value of π‘₯, the slopes of our function is increasing. This means it must be a local minima.

We can do exactly the same at our other critical point when π‘₯ is equal to one-half. We’ll write this as 0.5. We need to substitute 0.5 into our expression for 𝑓 double prime of π‘₯. We get negative 10 over three plus one divided by six times 0.5 squared. This time, if we evaluate this expression, we get negative eight divided by three, which we need to notice is negative. So when π‘₯ is equal to 0.5, the second derivative of 𝑓 of π‘₯ is negative. This means the slopes of our function are decreasing, which means we must be at a local maxima.

But remember, there’s one more thing we need to do. The question wants us to actually find the values of the local maxima and local minima of this function. To do this, all we need to do is substitute the values of our critical points into our function 𝑓 of π‘₯. So let’s clear some space and find the values of our local maxima and local minima. We’ll start when π‘₯ is equal to 0.1.

Substituting π‘₯ is equal to 0.1 into our function 𝑓 of π‘₯, we get negative five times 0.1 squared divided by three plus two times 0.1 minus one-sixth times the natural logarithm of 0.1. And if we evaluate this expression, we can simplify it to get 11 divided by 60 minus one-sixth times the natural logarithm of one over 10.

And we can do the same for our other critical point when π‘₯ is equal to one-half. We’ll write this as 0.5 and substitute π‘₯ is equal to 0.5 into our function 𝑓 of π‘₯. We get negative five multiplied by 0.5 squared divided by three plus two times 0.5 minus one-sixth times the natural logarithm of 0.5. Then, evaluating and simplifying this expression, we get seven divided by 12 minus one-sixth multiplied by the natural logarithm of one-half. And remember, we already showed that this is a local maxima of our function, and this gives us our final answer.

Therefore, we were able to show that the function 𝑓 of π‘₯ is equal to negative five π‘₯ squared over three plus two π‘₯ minus one-sixth times the natural logarithm of π‘₯ only has two local extrema. It has a local minimum with a value of 11 over 60 minus one-sixth times the natural logarithm of one over 10 when π‘₯ is equal to one-tenth. And it has a local maximum value of seven over 12 minus one-sixth times the natural logarithm of one-half when π‘₯ is equal to one-half.

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