### Video Transcript

Find, if any, the local maxima and local minima of the function π of π₯ is equal to π to the power of negative π₯ to the fourth power.

The question gives us a function π of π₯ which is an exponential function composed with a polynomial. And it wants us to find all of the local maxima and local minima of this function if any of these exist. The first thing we need to recall when weβre asked to find the local maxima and local minima of any function is that local extrema always occur at critical points of our function. And we say that π₯ is equal to π is a critical point of the function π of π₯ if π prime of π is equal to zero or π prime of π does not exist.

So, to find the local extrema, we need to find the critical points. And to do that, we need to find an expression for π prime of π₯. We see that π of π₯ is the composition of two functions; itβs the composition of an exponential function and a polynomial. So, to differentiate this, weβre going to need to use the chain rule. The chain rule tells us if we have π’ is a function of π£, and π£ in turn is a function of π₯, then the derivative of π’ of π£ of π₯ is equal to π£ prime of π₯ times π’ prime evaluated at π£ of π₯.

So, to use the chain rule on our function π of π₯, weβll set π£ to be our inner function negative π₯ to the fourth power. This gives us that π of π₯ is equal to π to the power of π£. So, now, we can see weβve rewritten π of π₯ to be a function of π£. To use the same notation in our chain rule, weβll set π’ to be π to the power of π£. So, by the chain rule, we have π prime of π₯ is equal to π£ prime of π₯ times π’ prime evaluated at π£ of π₯.

So, letβs find expressions for π’ prime of π₯ and π£ prime of π₯. Letβs start with π£ prime of π₯. Itβs the derivative of negative π₯ to the fourth power with respect to π₯. And we can do this by the power rule for differentiation. We multiply by the exponent of π₯, which is four, and reduce this exponent by one. We get π£ prime of π₯ is equal to negative four π₯ cubed.

Letβs now find an expression for π’ prime of π£. Thatβs the derivative of π to the power of π£ with respect to π£. In fact, we know the derivative of the exponential function is just equal to itself. So, π’ prime of π£ is also equal to π to the power of π£. Substituting in our expressions for π£ prime, π’ prime, and π£ of π₯, we get that π prime of π₯ is equal to negative four π₯ cubed times π to the power of negative π₯ to the fourth power.

And remember, weβre looking for the critical points of this function. So, weβre looking for values where our derivative does not exist or values where the derivative is equal to zero. In our case, we can see the domain of π prime of π₯ is all real numbers. It exists for any real value of π₯. So, our only critical points will be the values where our derivative is equal to zero. So, letβs solve π prime of π₯ is equal to zero.

We can see that π prime of π₯ is the product of two functions. And the product of two functions is equal to zero means that one of these two factors must be equal to zero. So, either negative four π₯ cubed is equal to zero or π to the power of negative π₯ to the fourth power is equal to zero. But π is a positive number, so π raised to the power of any value will be positive. So, π to the power of negative π₯ to the fourth power is positive for all values of π₯, so it canβt be equal to zero.

This means our only critical points of the function π of π₯ will be when negative four π₯ cubed is equal to zero. And, of course, the only solution to negative four π₯ cubed is equal to zero is when π₯ is equal to zero. So, our only possible value for a local extrema is when π₯ is equal to zero.

We now need to check what type of point this is. Weβll do this by using the first derivative test. To use the first derivative test, we want to evaluate the derivative of our function π around all of our critical points. Our only critical point was when π₯ is equal to zero. So, weβll check the derivative below zero at negative one and above zero at one. Remember, we already showed that π₯ is equal to zero was a critical point by showing π prime of zero is equal to zero. So, we can fill this value in our table if we want.

Letβs fill in the rest of the values of our table. Weβll start by finding π prime of negative one. Itβs equal to negative four times negative one cubed times π to the power of negative one times negative one to the power of four. Which we can simplify to give us four π to the power of negative one. And the information weβre interested in here is this is the product of two positive numbers. So, this is positive. So, the slope of our function is positive at π₯ is equal to negative one.

Letβs now do the same for π₯ is equal to one. We get π prime of one is equal to negative four times one cubed times π to the power of negative one times one to the fourth power. Which we can simplify to give us negative four times π to the power of negative one. This time, this is a negative number multiplied by a positive number, so this is less than zero. So, the slope of our function π is negative when π₯ is equal to one.

Letβs draw a quick sketch of what weβve shown. When π₯ is equal to negative one, our slope is positive. However, when π₯ is equal to zero, our slope is equal to zero. And when π₯ is equal to one, our slope is negative. In other words, when π₯ is equal to zero, we have a local maxima. And the last thing weβre gonna want to do is find the value of this local maxima. Weβll do this by substituting π₯ is equal to zero into our function π of π₯.

This gives us π to the power of negative one times zero to the fourth power. Thatβs π to the zeroth power, which is equal to one. Therefore, weβve shown the function π of π₯ is equal to π to the power of negative π₯ to the fourth power only has one local extrema. Itβs a local maximum at a value of one where π₯ is equal to zero.