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Find: (a) $f g(x)$ ) and; (b) $g(f(x))$.$$f(x)=\frac{x-2}{x+4}, \quad g(x)=\frac{1}{x-1}.$$

(a) $\frac{3-2 x}{4 x-3}$(b) $\frac{-(x+4)}{6}$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 6

The Chain Rule

Derivatives

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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.

02:57

Find a. $(f \circ g)(x)$, …

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finda. $(f \circ g)(x)…

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Finda. $(f \circ g)(x)…

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Find: (a) $f g(x)$ ) and; …

03:28

Find$$f(x)=x+4, g(…

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01:55

01:11

03:42

Find(a) $(f \circ g)(x…

03:06

01:45

01:53

0:00

Find (g.f) (x)

eso We're going to the composition function, and we have f is defined as X minus two or X plus four, and G is to find us one over X minus one. And, yeah, I think we're ready to just jump into the problem where we do f of g right, which is the function of and then what you're gonna do. So I'm just doing substitution replacing. And now when I do f of that, I just think about how, um, in my f function, I have something minus two over something. Plus four. Well, the X in my problems being replaced with this one over X minus one. So, everywhere I have an ex after replaced with one over X minus one, I'm one over X minus one. Now, I know several teachers will actually just let their students leave their answer like this. Um, but for the sake of getting the same do groups, we're supposed to circle everything. But if you got the same denominator for everything, you would have one minus, uh, you know, two times the quantity of X minus one all over one plus four times the quantity of X minus one and you can actually cancel out the X minus one because it's the same denominator on top and bottom. Otherwise, you can work it out and see that that is the same. So as you distribute, you would have negative to exit would become plus two plus one plus three and in the denominator before X minus four plus one and being minus three. And that is a simplified answer for part A. When we do part B, though, or doing G of F Becks, it's the same kind of work, you know. You start with ffx, which is that X minus two over X plus four. Remember the order of operations you do inside the parentheses first and then g of that. Well, I'm looking at G, uh, says kind of bubbly. We have one over something minus one. Well, now you replace that something this X, with X minus two over X plus four Again, some teachers will let you leave your answer like this. Other teachers are not as friendly where they'll make you rewrite it as experts to minus getting the same denominator. Uh, which is X plus four. And then as you divide, it's a multiplying by the reciprocal you'll get X Plus four into the numerator and then the denominator be one X minus one x zero X native. Two months for being negative. Six. And I would leave an answer like that. You can rewrite it. Otherwise so, um so Anyway, the best answer is this one and this one.

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