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3 years ago

If f and g are functions, then (fog)(x) is always equal to (gof)(x) true or false

Mathematics

1 answer:

valentina_108 [34]3 years ago

4 0

Answer:

False

Step-by-step explanation:

Here is a counter example:

● f(x) = x^2

● g(x) = 3x

● (f○g)(x) = (3x)^2 = 9x^2 (1)

● (g○f)(x) = 3(x^2) = 3x^2 (2)

We can see that (1) and (2) aren't equal to each others.

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Which of the following functions are homomorphisms?

Vikentia [17]

Part A:

Given $f:Z \rightarrow Z,$ defined by $f(x)=-x$

$f(x+y)=-(x+y)=-x-y \\ \\ f(x)+f(y)=-x+(-y)=-x-y$

but

$f(xy)=-xy \\ \\ f(x)\cdot f(y)=-x\cdot-y=xy$

Since, f(xy) ≠ f(x)f(y)

Therefore, the function is not a homomorphism.

Part B:

Given $f:Z_2 \rightarrow Z_2,$ defined by $f(x)=-x$

Note that in $Z_2$ , -1 = 1 and f(0) = 0 and f(1) = -1 = 1, so we can also use the formular $f(x)=x$

$f(x+y)=x+y \\ \\ f(x)+f(y)=x+y$

and

$f(xy)=xy \\ \\ f(x)\cdot f(y)=xy$

Therefore, the function is a homomorphism.

Part C:

Given $g:Q\rightarrow Q$ , defined by $g(x)= \frac{1}{x^2+1}$

$g(x+y)= \frac{1}{(x+y)^2+1} = \frac{1}{x^2+2xy+y^2+1} \\ \\ g(x)+g(y)= \frac{1}{x^2+1} + \frac{1}{y^2+1} = \frac{y^2+1+x^2+1}{(x^2+1)(y^2+1)} = \frac{x^2+y^2+2}{x^2y^2+x^2+y^2+1}$

Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.

Part D:

Given $h:R\rightarrow M(R)$ , defined by $h(a)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)$

$h(a+b)= \left(\begin{array}{cc}-(a+b)&0\\a+b&0\end{array}\right)= \left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right) \\ \\ h(a)+h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)+ \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)=\left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right)$

but

$h(ab)= \left(\begin{array}{cc}-ab&0\\ab&0\end{array}\right) \\ \\ h(a)\cdot h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)\cdot \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)= \left(\begin{array}{cc}ab&0\\-ab&0\end{array}\right)$

Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.

Part E:

Given $f:Z_{12}\rightarrow Z_4$ , defined by $\left([x_{12}]\right)=[x_4]$ , where $[u_n]$ denotes the lass of the integer $u$ in $Z_n$ .

Then, for any $[a_{12}],[b_{12}]\in Z_{12}$ , we have

$f\left([a_{12}]+[b_{12}]\right)=f\left([a+b]_{12}\right) \\ \\ =[a+b]_4=[a]_4+[b]_4=f\left([a]_{12}\right)+f\left([b]_{12}\right)$

and

$f\left([a_{12}][b_{12}]\right)=f\left([ab]_{12}\right) \\ \\ =[ab]_4=[a]_4[b]_4=f\left([a]_{12}\right)f\left([b]_{12}\right)$

Therefore, the function is a homomorphism.

7 0

3 years ago

Which of the following are quadratic functions?<br> Check all that apply.

Elanso [62]

B and c, anything with an x^2 term is a quadratic.

8 0

4 years ago

Read 2 more answers

Solve y = x - 5 if the domain is -3

DedPeter [7]

Answer:

y = -8

Step-by-step explanation:

The domain is the x values so x=-3

y = x-5

Substituting x=-3

y = -3-5

y=-8

6 0

3 years ago

I REALLY NEED HELP, PLEASE HELP !! WORTH 30 POINTS !!

valina [46]

Answer:

2/9

Step-by-step explanation:

y=6(3)^x

Let x =-3

y = 6 * 3^-3

Remember negative exponents move it from the numerator to the denominator

y = 6 *1/3^3

y = 6 * 1/27

y = 6/27

y = 2/9

8 0

4 years ago

Us<br> Find the area shaded in yellow.<br> [ ? ] square units

ira [324]

Answer:

45 square units

Step-by-step explanation:

You can cut the yellow figure into a triangle (4 by 5) and a rectangle (7 by 5).

Using the formula $\frac{1}{2} *b*h$ and $l*w$ area formulas, we get:

$\frac{1}{2} *4*5+7*5=\\2*5+7*5=\\10+35=\\45$

The area shaded in yellow is 45 square units

6 0

3 years ago