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Use the first derivative to determine where the given function is increasing and decreasing.$r(t)=4 t^{3}-15 t^{2}+18 t+2$

Dec on $1 < x < 3 / 2 ;$ inc on $x < 1$ or $x > 3 / 2$

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 2

The First Derivative Test

Derivatives

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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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Yeah, this is chapter three, Section two, Problem eight. In this problem, we are asked to find where our function R of T is increasing and where it is decreasing by finding the first derivative. So the first step is finding the critical points. And if you remember, the critical points are where the derivative of our function is equal to zero. So I already went ahead and found the derivative. Our function, which I got to be our prime of T is equal to 12 t squared minus 30 t plus 18. I took six out of that to make it the problem a little easier. And I got six times to t squared minus five t plus three. That second part factor down even more to get six times to t minus three times t minus one. And that made it really easy to find are critical points. Where are our prime of T would equal zero and that would be at T equals three halfs and it t equals one. So from here we can draw our number line. So here we're going to have our t equals one, and here we're going to have our t equals three halves. And we know of both of these points were hot zeros. So what we're gonna do is we're going to pick a number that's less than one. We're going to pick a number that's in between one and one half or 3/2. And we're going to pick a number that is greater than three halves. And we're going to plug in each of those numbers into our prime of T to see whether our function is increasing or decreasing. So, for down here, let's choose T equals zero. Yeah, for our tea. In between one and 1.5, we can do one in 1/4 which is also going to be equal to t equals 5/4 and just over t equals three halfs. We can do t equals two. Yeah, So we're gonna plug each of these numbers into our prime of T and see what we get. So first, let's do our prime of T equals zero. Yeah, we're gonna get six times zero minus three times zero minus one. That's gonna give us six times negative three times negative one, and that's going to give us a positive 18. So if we have a T that's less than one are function is going to be increasing. Mhm. No. Let's plug in 5/4. Yeah, our prime of 5/4 or 1.25 Mhm. It's gonna be five halves minus three times 5/4 minus one. Yeah, that's gonna give us six times negative one half times one fourth. And that's going to give us negative 3/4. So because that is negative, we know that between two equals one and two equals one half. Our function is going to be decreasing now. Lastly, let's plug in T equals two. So our prime of T equals two. That's going to give us six times four minus three times two minus one. It's gonna give us six times one times one, which gives us a positive six. And because that's positive, we know that if we have a T value greater than three over to our function will be increasing Holmes. So to sum it all up for our function R of T, it will be increasing if t is less than one or if t is greater then 3/2 and our function will be decreasing. Yeah, if t is in between one. Yeah, and 3/2. Oh, yeah. Yeah. And this is our answer

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