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Homework: HW2Score: -28 0l 42 (42 completel1.6.3Use Ihe Product Rula (llnd Ine derivativa (Iho Gvum funclion Find dun OLNvc oxpanding Iha produc} Iirsl((x) = (- 61(...

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Homework: HW2Score: -28 0l 42 (42 completel1.6.3Use Ihe Product Rula (llnd Ine derivativa (Iho Gvum funclion Find dun OLNvc oxpanding Iha produc} Iirsl((x) = (- 61(42 + 1)Use the codud {ulefind Iha: ccrivalivo of the tunction . Selectthe corec: ans re DelctancTheand2completeTho dorivativa6. 1) ( 6MAx + 1) -ine dencelndUne den ulnX(x + 1)Una duni utrat(x- G12 GMAx + 1)(donvabtcho c

Homework: HW2 Score: - 28 0l 42 (42 completel 1.6.3 Use Ihe Product Rula (llnd Ine derivativa (Iho Gvum funclion Find dun OLNvc oxpanding Iha produc} Iirsl ((x) = (- 61(42 + 1) Use the codud {ule find Iha: ccrivalivo of the tunction . Selectthe corec: ans re Delctanc Theand2 complete Tho dorivativa 6. 1) ( 6MAx + 1) - ine dencelnd Une den uln X(x + 1) Una duni utrat (x- G12 GMAx + 1)( donvabt cho c



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A computer was used to complete the preliminary calculations; form the extensions table; calculate the summations
$\Sigma x, \Sigma y, \Sigma x^{2}, \Sigma x y, \Sigma y^{2}$; and find the $\operatorname{SS}(x), \operatorname{SS}(y),$ and $\operatorname{SS}(x y)$ for the following set of bivariate data. Verify the results by calculating the values yourself. $$\begin{array}{lllllll} \hline x & 45 & 52 & 49 & 60 & 67 & 61 \\ y & 22 & 26 & 21 & 28 & 33 & 32 \\ \hline \end{array}$$

Okay, so first I want to compute the composite function F g of X. Remember, a composite is when you're putting a one function inside of another. So I'm going to replace G of X, which is four x plus one inside of FX, replaced with X. So this will be 1/4 for X plus one minus one. Simplify what's in the prince's first one in one cancel. So I get 1/4 times for X 1/4 times four is just one X Okay. And then let's try the other composite function. I'm plugging in ffx now this whole function in for X of the function G. So this will be four times 1/4 times X minus one plus one. So first things first. Well, actually, his four isn't distributed to this entire term. So four and 1/4 this will just be ones that will be X minus one plus one. Those will cancel. I'll just get X. So algebraic Lee. I could see that f of G of X is equal to G of f of X since they both equal x or one X So let's figure these out on the is most graphing calculator. So if I photographed 1/4 that's what's one. And then g of X is for X. Look, this medicine that's one now to compute f of G f X and then g of f of X. I will see here that these two will also be the graph. That's so they're all the same function. So therefore, this proves that these two composites are equal to each other.

Alrighty. So first, let's find the function F of G of X. So what is the composite function? That's what I'm pulled. Um, putting one entire functioned in for X of another function. So I've just playing in X squared minus five plus six inside of the square root simplifying under the square root. This will be X squared. Plus what now? What is the domain of F of G of X? Well, this is when, um what is the definition of domain? It's all the X values that we can plug into a function. So looking at G of X g of X is just a polynomial. So my domain is all real numbers. What about in ffx? Well, I can't have what's inside the square root. It has to be greater than or equal to zero. So I'm gonna set what's inside the square. Greater dinner equal to zero. Subtract one. When I subtract when I keep the same sign, something squared is greater than negative number. Well, this is always going to be true. So therefore my domain will be total numbers. Let's try now the opposite function G of F FX, the opposite composites. Now I'm plugging in all of ffx into G. So this will be the square root of X plus six. Race to the second power minus five. Um, the square and the exponent will cancel, and this will not leave an absolute value, since the exponents on the outside X plus six minus five will end up being just X plus one. So are these two values equal X plus one equals the square root of X squared plus one. I'm going to say no, they're not equal. And we'll show this on our graph. So we're going to graph on Dez most F of X is the square root of X plus six and M g of X is X squared minus five. Right. So turn off these graphs. I'm just gonna grab F g of X is the Green Line like so in? This shows that that the main is again all real numbers I can plug in any value on G of F. FX is a completely different line, right? These are two different functions. So therefore, these two opposite composite functions are not equivalent

Already. So first I'm gonna find a composite f of G of X. So in this case, I'm replacing. When I found the composite, I'm plugging in a function into another function. So I'm plugging in to x the third into, um ffx. So this will be absolute value of two X to the third. Okay, Now let's try theocracy it composite function g of f of X. So I'm replacing this entire function inside of G of X of this will be to times absolute value of X, all raised to the third power so I could simplify these absolute values, right? This is the same as two times the absolute value of X the third, which I know that anything even could be pulled out right. And this is the same is X squared times x the first. So this is two X squared times absolute value of X and this explain it can come inside the absolute value and similarly, two times x squared times the absolute value of X. So here I can prove that these two functions thes two composite functions are the same. Let's try this on Dismas. So let's say I graph ffx is the absolute value. Quickly. Absolute value of X and D of X is two X to the third. Okay, And then I'm gonna graph f of G of X And let's see if g f fx produce the same. So they fall right on top of each other, and if we zoom out even more, you'll see that they're still the same function. So therefore, this proves that these two composites are equal.

In this question we will learn about the basic question of the linear equation. Like so we have a function effects which is it was too three x. last seven we have to find the jewel of the given function, which means that we have to find the value of X. For which the given function values to the like. So to find the values of X. We can substitute effects. It's equals to zero. So they so three X plus seven is a close to judo Or three X is equal to -7. So the value of tax would be -7 x three. Right? So for the value of x equals 2 -7 x three. The function value is zero. Like so this is the judo of the given function F X equals 23 X plus seven. So in this way we can find the genitals of any function. By substituting by creating the function value to judo. We can find the value of X. Right? Yes.


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