Question

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 (f*g)(x). Assume all appropriate restrictions to the domain.
a) (f*g)(x) = x^2+5x-14
b) (f*g)(x) = x^2+9x-14
c) (f*g)(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, ∞)

Answer 1

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 $g(x)=3$.

Then, $(f+g)(x)=(x+4)^2+3$.

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]

Answer 2

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.