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Use the first derivative to determine where the given function is increasing and decreasing.$f(x)=3 x^{2}-2 x+1$

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

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 2

The First Derivative Test

Derivatives

Campbell University

Oregon State University

University of Michigan - Ann Arbor

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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This is chapter three, Section two, Problem three. And this problem is asking us to use the first derivative to find where this function F of X is increasing and where it is decreasing. So the first thing that we're going to do is find are critical points. And if you remember, the critical points are where the derivative of the function is equal to zero. So in this case, where F prime of X is equal to zero, So the first thing we're going to do is find the derivative of F of X and find those critical points so f prime of X. It looks like we're going to have two times three is six X minus two factored out. We can also write this as two times three X minus one. So f prime of X or what we have right here are critical points are where this is equal to zero. So if we set this whole thing to zero or just this part, we can set that equal to zero and solving for X. We get a critical point of X equals one third. From here, we can create our number line with our critical point here in the center, we can put one third and we know if we plug in one third into our F prime of X, we're going to get a value of zero. So for our other points, we're going to pick any arbitrary number one that is less than one third and one that is greater than one third for simplicity. Let's choose zero and one. So if we have ex prime ex prime of zero, yeah, that's going to give us two times three times zero minus one. That's going to be equal to two times negative one or negative to. That means that if we have an ex of less than one third, our function is going to be decreasing. My let's try one. If we have F prime of wine, that's going to give us two times three times one minus one. That's two times two or four, which is positive, which means that any X value over one third is going to give us a positive value and our function is going to be increasing. So now from this we can say that our function is going to be decreasing if X is less than one third and F of X is going to be increasing if X is greater than one third mhm

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