Mathematics, 31.01.2020 07:52, eboniwiley

1. let f(x) = 4x-3/x-10 and g(x) = 2x-8/x-10. find (f+g)(x). assume all appropriate restrictions to the domain.
a) (f+g)(x) = 6x-11/x-10
b) (f+g)(x) = 2x+5/x-10
c) (f+g)(x) = 6x+11/x-10
d) (f+g)(x) = 6x-11/2x-20
2. find the domain of the function (f/g)(x) where f(x)=x²-9 and g(x)=x²-4x+3.
a) (-∞,∞)
b) (-∞,1) u (1, ∞)
c) (-∞,1) u (1,3) u (3, ∞)
d) (-∞,-3) u (-3,-1) u (-1, ∞)
3. let f(x) = √x-2 and g(x) = √x+7. find (fg)(x). assume all appropriate restrictions to the domain.
a) (fg)(x) = x^2+5x-14
b) (fg)(x) = x^2+9x-14
c) (fg)(x) = √x²+9x-14
d) (f*g)(x) = √x²+5x-14
4. determine the graph of (f-g)(x) when f(x) = 1/x and g(x) = √x.
5. find the range of (f+g)(x) when f(x)=(x+4)² and g(x)=3.
a) (-∞,∞)
b) [3, ∞)
c) (-∞, -3)
d) [-3, ∞)

Answers: 1

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Answers

Part 1.

Given that $f(x) = \frac{4x-3}{x-10}$ and $g(x) = \frac{2x-8}{x-10}$

Then, $(f+g)(x)=\frac{4x-3}{x-10}+\frac{2x-8}{x-10}=\frac{4x-3+2x-8}{x-10}=\frac{6x-11}{x-10}$ .

Therefore, the correct anser is option a

Part 2.

Given f(x)=x²-9 and g(x)=x²-4x+3.

Then, $\left(\frac{f}{g}\right)(x)= \frac{x^2-9}{x^2-4x+3} = \frac{(x-3)(x+3)}{(x-1)(x-3)} =\frac{x+3}{x-1}$

The domain of $\left(\frac{f}{g}\right)(x)$ is all real numbers except for the value of x for which the denominator is 0.

i.e. x - 1 = 0 implies that x = 1.

Therefore, the domain of $\left(\frac{f}{g}\right)(x) is (-∞,1) U (1, ∞)$ [option b]

Part 3.

Given that $f(x) = \sqrt{x-2}$ and $g(x) = \sqrt{x+7}$ .

Then
$(f*g)(x)=\sqrt{x-2}*\sqrt{x+7} \\ \\ =\sqrt{(x-2)(x+7)}=\sqrt{x^2+5x-14$
[option d]

Part 4.
Given that $f(x) = \frac{1}{x}$ and $g(x) = \sqrt{x}$ .

Then, $(f-g)(x)=\frac{1}{x}-\sqrt{x}$

The graph of $(f-g)(x)=\frac{1}{x}-\sqrt{x}$ is attached.

Part 5.

Given that f(x)=(x+4)^2

and

.

Then,

.

The vertex of (f + g)(x) is (-4, 3),
The range of (f+g)(x)=(x+4)^2+3

is all real numbers greater than or equal to 3.

Therefore, the reange of (f + g)(x) is [3, ∞) [option b]

1. let f(x) = 4x-3/x-10 and g(x) = 2x-8/x-10. find (f+g)(x). assume all appropriate restrictions to

1. let f(x) = 4x-3/x-10 and g(x) = 2x-8/x-10. find (f+g)(x). assume all appropriate restrictions to

answered: clairajogriggsk

The correct answers are:

(1) Option (a) $(f+g)(x) = \frac{6x-11}{x-10}$

(2) Option (b) $(-\infty, 1) ~\bigcup ~(1, +\infty)$

(3) Option (d) $(f*g)(x) = \sqrt{x^2 + 5x - 14}$

(4) Graph is attached with the answer along with the explanation (below)!

(5) Option (b) $[3, \infty)$

Explanations:

(1) Given Data:

f(x) = $\frac{4x-3}{x-10}$

g(x) = $\frac{2x-8}{x-10}$

Required = (f+g)(x) = ?

The expression (f+g)(x) is nothing but the addition of f(x) and g(x). Therefore, in order to find (f+g)(x), we need to add both the given functions as follows:

$(f+g)(x) = \frac{4x-3}{x-10} + \frac{2x-8}{x-10}$

Now we need to simplify the above equation as follows:

$(f+g)(x) = \frac{4x-3 + 2x - 8}{x-10} \\ (f+g)(x) = \frac{6x-11}{x-10}$

Hence the correct answer is $(f+g)(x) = \frac{6x-11}{x-10}$ Option (a)

(2) Given Data:

f(x) = x^2 - 9

g(x) = x^2 - 4x + 3

Before finding the domain of the expression $(\frac{f}{g})(x)$ , we need to first evalute that expression as follows:

$(\frac{f}{g})(x) = \frac{f(x)}{g(x)} \\ Plug~in~the~values~of~f(x)~and~g(x)~in~the~above~equation.\\ (\frac{f}{g})(x) = \frac{x^2-9}{x^2-4x+3} \\ (\frac{f}{g})(x) = \frac{(x-3)(x+3)}{(x-3)(x-1)} \\ (\frac{f}{g})(x) = \frac{(x+3)}{(x-1)}$

Now we need to put the denominator equal to zero in order to know what values of x should not be in the domain of this function:

x-1 = 0

x = 1

It means that the domain of $(\frac{f}{g})(x)$ is all real numbers EXCEPT x = 1. The (closed) parentheses " ) " or "(" means that the number is not included in the domain. Therefore, we can write that the domain of $(\frac{f}{g})(x)$ is $(-\infty, 1) ~\bigcup ~(1, +\infty)$ (Option b)

(3) Given Data:

f(x) = $\sqrt{x-2}$

g(x) = $\sqrt{x+7}$

Required = (f*g)(x) = ?

The expression (f*g)(x) is nothing but the multiplication of f(x) and g(x). Therefore, in order to find (f*g)(x), we need to multiply both the given functions as follows:

$(f*g)(x) = \sqrt{x-2} * \sqrt{x+7}$

Now we need to simplify the above equation as follows:

$(f*g)(x) = \sqrt{x-2} * \sqrt{x+7} \\ (f*g)(x) = \sqrt{(x-2)(x+7)} \\ (f*g)(x) = \sqrt{x^2 + 7x -2x - 14}\\ (f*g)(x) = \sqrt{x^2 + 5x - 14}$ (Option d)

(4) Given Data:

f(x) = $\frac{1}{x}$

g(x) = $\sqrt{x}$

Required = The graph of (f-g)(x) = ?

Before plotting the graph let us evalute it first. (f-g)(x) is the subtraction of g(x) from f(x). Mathematically, we can write it as:

(f-g)(x) = $\frac{1}{x} - \sqrt{x}$

Now simplify:

$(f-g)(x) = \frac{1}{x} - \sqrt{x} \\ (f-g)(x) = \frac{1-x\sqrt{x}}{x} \\$

Look at the graph attached with this answer. As you can see, at x=0, the graph shoots up! As at x=0, the value of function approaches to infinity.

(5) Given Data:

f(x) = (x+4)^2

g(x) = 3

Required = Range of (f+g)(x) = ?

Before finding the range of (f+g)(x), we first need to write the function:

(f+g)(x) = (x+4)^2 + 3

Now that we have written the function, the next step is to find the inverse of this function in order to obtain the range.To find the inverse, swap x with y, and y with x and put (f+g)(x) = y as follows:

(f+g)(x) = y = (x+4)^2 + 3

Now swap:

x = (y+4)^2 + 3

Now solve for y:

(x-3) = (y+4)^2

Take square-root on both sides:

$\sqrt{(x-3)} = y+4$

$y = \sqrt{(x-3)} - 4$

As you know that the square root of negative numbers are the complex numbers, and in range, we do not include the complex numbers. Therefore, the values of x should be greater or equal to 3 to have the square-roots to be the real numbers. Therefore,

Range of (f+g)(x) = $[3, \infty)$ (Option b)

Note: "[" or "]" bracket is used to INCLUDE the value. It means that 3 is included in the range.

answered: Guest

a. 74 degrees

step-by-step explanation:

because the lines through the side and the box in the corners shows that these are congruent triangles and since they are mirrored the 37 degrees is also mirrored so the answer is 37 * 2 which is 74.