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Classify all critical points.$s(t)=(t-1)^{4}(t+3)^{3}$

$$\mathrm{M}(-9 / 7,137.511), \mathrm{m}(1,0), \mathrm{N}(-3,0)$$

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 2

The First Derivative Test

Derivatives

Harvey Mudd College

University of Nottingham

Idaho State University

Boston College

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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Classify all critical poin…

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Locate all critical points…

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Find all critical points o…

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Locate and classify all th…

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find first the first derivative of this function. So to do that we're going to do a combination of the exponents and the and the product rule. So that's basically a fancy or those are the two parts of the chain rule that we will be using. So first off we're gonna work with this expression right here and we're going to use the exponent rule and the chain rule. So bring down the floor T minus one. Subtract one from the exponents and then take the derivative of the inside was just one next. Or and then we're going to multiply that by this expression because that will be the product rule. So T-plus three to the third. And then we're going to add the derivative of this expression. So bring down the three from the explain it. You got two plus three squared times The derivative of the inside which is one. And then multiplied by This whole expression right here. So T -1 to the fourth. And we're going to do is set that whole equation 20 because we're trying to find the zeros of the derivative. So using this, there are two ways we can find the zeros. We have two products or the sum of two products here. So if any of those were two equals zero, or if any of these individuals were to equal zero and to equal the other part that would give us um a zero. So we want to solve this for T. So we'll start by doing that. I'm going to cut out some of the algebraic stabs because you should be able to figure them out on your own. But let me rewrite this equals, we're going to move this to this side negative three T plus three squared Times 2 -1, 4. So all I did so far was cancel out the um the exponents um to get each of them on their on their respective sides. So then solving for those we can just move things around to get to get our T. S. On either side and then multiply them out. So if you multiply out your teas first and then you get T. Equals yeah 97 So that is 10. Additionally if you were to look back at the original equation or the original derivative of for T -13 T. Plus three you'll notice that Both T -1 -1 And T. Plus three appear on in both of the products. So if either of those were two equals zero then you would have your whole function equal to zero. So we can solve for two plus three equals zero. As well as t minus one equals zero. Which will give us bring him back here -3 & one. So next we're going to plug these values back into our original equation of t minus one. Product to the fourth Power times, T plus three product to the three power. And when you plug it back in you will get your function values oh 1 37 About 1 37. And then zero and 0. Finally, what we want to do is put all of these values on a scale and pick value slightly above and slightly below them. So we have one negative 9/7 and negative three. And for value slightly above and slightly below, let's pick negative two, zero two and we'll do negative for over here. So we want to figure out if the derivative is increasing or decreasing at each of these values, we know at the Zeros, the derivative will equal zero. But then uh the other value slightly above and slightly below. You want to plug it into your derivative equation And see if the values are positive or negative. So that negative four um you can eyeball it or you can figure actually solve it out. I'm going to eyeball it just for the sake of time and simplicity. So at negative four we have a positive value. And then at native to we have a positive value as well at zero, we have a negative value and two we have a positive value. So at this We know that our ordered pair or are zeros of -30. This is an inflection point. Since you have the same sign on either side At -9 7th at at -97137. We have a maximum Since it's positive on the on the left and negative on the right. And then at 10 we have a minimum since it's negative on one side and positive on the other.

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