### Video Transcript

Determine the local maximum and minimum values of the function π of π₯ is equal to negative three times the natural logarithm of two π₯ squared plus three.

The question gives us a function π of π₯. And it wants us to find all the local maximum and minimum values of this function. The first thing we need to remember is local extrema will always occur at critical points of our function. And we recall that we say that itβs a critical point of our function π of π₯ when π₯ is equal to π if the derivative of π evaluated at π is equal to zero or the derivative does not exist at this point.

So to find our local extrema, we want to find the critical points of our function. And to do that, weβre going to need to find an expression for π prime of π₯. So weβre going to want to differentiate negative three times the natural logarithm of two π₯ squared plus three. And we can see that this is the composition of two functions. Weβre taking the natural logarithm of a quadratic.

And we know to differentiate the composition of two functions, weβll want to use the chain rule. We recall the chain rule tells us if π’ is a function of π£ and π£ is turn is a function of π₯, then the derivative of π’ composed with π£ of π₯ is equal to π£ prime of π₯ times π’ prime evaluated at π£ of π₯.

So to apply the chain rule to our function π of π₯, weβll start by setting π£ of π₯ to be our inner function. Thatβs two π₯ squared plus three. And then by using this definition of π£ of π₯, weβve rewritten π of π₯ to be negative three times the natural logarithm of π£. Then to keep our notation consistent with the chain rule, weβll set our function π’ of π£ to be negative three times the natural logarithm of π£.

Using these definitions of π’ and π£, we can see that π of π₯ is equal to π’ composed with π£. π’ is a function of π£, and π£ in turn is a function of π₯. This means we can now apply the chain rule to find π prime of π₯. Itβs equal to π£ prime of π₯ times π’ prime evaluated at π£ of π₯.

Now, to use the chain rule, weβre going to need to find expressions for π£ prime and π’ prime. Letβs start with π£ prime of π₯. Thatβs the derivative of two π₯ squared plus three with respect to π₯. And we can evaluate this by using the power rule for differentiation. We get π£ prime of π₯ is equal to four π₯.

We now want to find an expression for π’ prime of π£. Thatβs the derivative of negative three times the natural logarithm of π£ with respect to π£. And we know the derivative of the natural logarithm of π£ with respect to π£ is just equal to one over π£. So π’ prime of π£ is equal to negative three divided by π£.

Substituting in our expressions for π’ prime, π£ prime, and π£ of π₯, we get that π prime of π₯ is equal to four π₯ times negative three divided by two π₯ squared plus three. And finally, weβll simplify this expression. We get that π prime of π₯ is equal to negative 12π₯ divided by two π₯ squared plus three. Remember, we want to find the critical points of our function π of π₯. Thatβs where the derivative is equal to zero or where the derivative does not exist.

Letβs start by finding all the points where our derivative does not exist. We can see that π prime of π₯ is a rational function. So our rational function will be defined for all values of π₯ except where the denominator is equal to zero. But if we were to try and solve our denominator is equal to zero, we would get that π₯ squared is equal to negative three over two. But we know that π₯ squared is greater than or equal to zero for all real values of π₯. So there are no values of π₯ which will make our denominator equal to zero. This means the only possible critical points for this function is where the derivative is equal to zero.

So letβs try and find the values of π₯ where our derivative is equal to zero. Again, since this is a rational function, for this function to be equal to zero, our numerator must be equal to zero. And if 12π₯ is equal to zero, then we just have π₯ is equal to zero. And of course, normally, we would check that our denominator is not equal to zero when π₯ is equal to zero. However, weβve already shown that this is true. This means weβve shown that there is a critical point when π₯ is equal to zero for our function π of π₯. In fact, weβve shown this is the only critical point for our function π of π₯.

But weβre not done yet. Remember, weβre looking for the local maxima and local minima of our function. So we need to check if this critical point is either a local maxima or a local minima or neither. Since weβve already found an expression for π prime of π₯ and π prime of π₯ is defined for all of our values of π₯, weβll do this by using the first derivative test.

Since our only critical point is when π₯ is equal to zero, weβll want to check the slope of our function above and below this value. Weβll check π₯ is equal to negative one and π₯ is equal to one. So letβs start filling out our table.

First, we already know that π prime of zero is equal to zero. Next, letβs find the slope of our function when π₯ is equal to negative one. Weβll substitute π₯ is equal to negative one into our expression for π prime of π₯. This gives us negative 12 times negative one divided by two times negative one squared plus three. And if we calculate this expression, we get 12 divided by five. So weβve shown that π prime of π₯ is positive when π₯ is equal to negative one.

We can do the same to find our derivative when π₯ is equal to one. We get negative 12 times one divided by two times one squared plus three. And if we calculate this, we get negative 12 divided by five, which we know is negative.

So letβs sketch what weβve shown. Weβve shown when π₯ is equal to negative one, our slope is positive. When π₯ is equal to zero, our slope is equal to zero. And when π₯ is equal to one, our slope is negative. And from the sketch, we can see when π₯ is equal to zero, weβll have a local maximum.

The last thing we want to do is find the value of this local maximum. Weβll do this by substituting π₯ is equal to zero into our function π of π₯. This gives us negative three times the natural logarithm of two times zero squared plus three. And we can then evaluate this. We get negative three times the natural logarithm of three.

Therefore, we were able to show the function π of π₯ is equal to negative three times the natural logarithm of two π₯ squared plus three only has one local extrema. Which is the local maximum of value negative three times the natural logarithm of three at π₯ is equal to zero.