Find the critical points of the function and classify them as local maxima or local minima or ne

Find the critical points of the function and classify them...

Key Concepts

First Derivative Test

The first derivative test is used to determine whether a critical point is a local maximum, local minimum, or neither. By analyzing the change in the sign of the derivative before and after the critical point, one can classify it as a maximum (if the derivative changes from positive to negative), a minimum (if the derivative changes from negative to positive), or neither if no such sign change occurs.

Second Derivative Test

The second derivative test provides another method for classifying the nature of a critical point by evaluating the concavity of the function at that point. A positive second derivative indicates concavity upward and a local minimum, while a negative second derivative indicates concavity downward and a local maximum. If the second derivative is zero, the test is inconclusive, requiring further analysis.

Critical Points

Critical points are points in the domain of a function where its derivative is zero or does not exist. These points are significant because they are candidates for local extrema or saddle points, and identifying them helps in understanding the overall behavior of the function.

Product Rule

The product rule is a fundamental differentiation technique used when computing the derivative of a product of two functions. It states that the derivative of the product is equal to the derivative of the first function multiplied by the second function plus the first function multiplied by the derivative of the second function.

Transcript

00:01 Hey everyone, so for this problem, we're even a function and asked to find the critical points of that function, and then figure out if they're local maxima or local minima.

00:10 And so the first thing we want to do is find the critical points.

00:13 The way we do that, if we want to see where the derivative of the function is equal to zero, that's going to give us those critical points.

00:23 And so first we need to find the derivative.

00:27 For this, we're going to have to use the product rule.

00:31 The derivative of the first part times the second part, plus the derivative of the second part, which is negative 3, e to the negative 3x, times the first part.

00:45 Okay, and so we're going to condense that a little bit more.

00:49 We have e to the negative 3x minus 3x, the same of 3x.

01:00 And so we're going to factor this out just one more step before we set it equal to 0.

01:07 We're going to factor out that e to the negative 3x, and this is what we get.

01:12 And so now we're going to set that equal to zero and try to solve for x.

01:16 Now this e function, no matter what you plug into this x, e cannot be zero.

01:24 Right? it's not defined for zero.

01:27 So the only way for this whole thing to be zero is if the second part is zero.

01:36 Okay, so we're just going to set that second part equal to zero.

01:40 Solve it and we get x equals one third awesome so that is our critical point that is where there's some big change in the graph of maximum minimum or neither which means there's a change in concavity and so figure out what kind of critical point this is if it's a local max a local men or neither there's two ways to do this and we're going to do it both ways.

02:15 So the first way is using the first derivative test.

02:18 So the first derivative test, we draw a number line and we put our critical points on it.

02:26 So we only have one critical point.

02:28 It's x equals one third.

02:30 And this is where our derivative is equal to zero, right? but our first derivative test tells us that wherever the derivative is negative, where it's less than then we're going to be decreasing.

02:50 And wherever our derivative is positive, the original function is increasing.

02:57 Right? so all we have to do is test points on either side to see if it is a negative value or a positive value, and that'll tell us if it's increasing or decreasing on that interval.

03:10 Because if this is the only place where it changes direction, then testing one point over here will tell us what is doing.

03:17 For the whole area.

03:20 So let's do that.

03:25 Let's try 0, and we'll test 1.

03:32 We want to pick easy points because any point will work, so let's just use the ones that are easiest.

03:37 And we're going to plug that into our derivative.

03:40 So f prime of 0, remember it's e to the negative 3, plug in our 0, times 1 minus 3, and plug in our 0 again.

03:51 So this is going to give us, so negative 3 times 0 is 0, right? e to the 0 times 1 minus 0, that's 1 times 1, and we get 1.

04:02 And so that's nice that it's an easy number, but we don't really care that it's an easy number.

04:07 We care that it's positive, right? that means that we're increasing on this interval...

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