The correct answers are:
(1) Option (a) [tex](f+g)(x) = \frac{6x-11}{x-10} [/tex]
(2) Option (b) [tex] (-\infty, 1) ~\bigcup ~(1, +\infty) [/tex]
(3) Option (d) [tex] (f*g)(x) = \sqrt{x^2 + 5x - 14} [/tex]
(4) Graph is attached with the answer along with the explanation (below)!
(5) Option (b) [tex] [3, \infty)[/tex]
Explanations:
(1) Given Data:
f(x) = [tex] \frac{4x-3}{x-10} [/tex]
g(x) = [tex] \frac{2x-8}{x-10} [/tex]
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:
[tex] (f+g)(x) = \frac{4x-3}{x-10} + \frac{2x-8}{x-10} [/tex]
Now we need to simplify the above equation as follows:
[tex] (f+g)(x) = \frac{4x-3 + 2x - 8}{x-10} \\
(f+g)(x) = \frac{6x-11}{x-10} [/tex]
Hence the correct answer is [tex](f+g)(x) = \frac{6x-11}{x-10} [/tex] Option (a)
(2) Given Data:
f(x) = [tex]x^2 - 9[/tex]
g(x) = [tex]x^2 - 4x + 3 [/tex]
Before finding the domain of the expression [tex] (\frac{f}{g})(x) [/tex], we need to first evalute that expression as follows:
[tex] (\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)}
[/tex]
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 [tex] (\frac{f}{g})(x) [/tex] 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 [tex] (\frac{f}{g})(x) [/tex] is [tex] (-\infty, 1) ~\bigcup ~(1, +\infty) [/tex] (Option b)
(3) Given Data:
f(x) = [tex]\sqrt{x-2}[/tex]
g(x) = [tex]\sqrt{x+7}[/tex]
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:
[tex] (f*g)(x) = \sqrt{x-2} * \sqrt{x+7}[/tex]
Now we need to simplify the above equation as follows:
[tex] (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} [/tex] (Option d)
(4) Given Data:
f(x) = [tex] \frac{1}{x} [/tex]
g(x) = [tex] \sqrt{x} [/tex]
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) = [tex] \frac{1}{x} - \sqrt{x} [/tex]
Now simplify:
[tex] (f-g)(x) = \frac{1}{x} - \sqrt{x} \\
(f-g)(x) = \frac{1-x\sqrt{x}}{x} \\
[/tex]
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) = [tex] (x+4)^2 [/tex]
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) = [tex] (x+4)^2 + 3 [/tex]
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 = [tex] (x+4)^2 + 3 [/tex]
Now swap:
x = [tex] (y+4)^2 + 3 [/tex]
Now solve for y:
[tex] (x-3) = (y+4)^2 [/tex]
Take square-root on both sides:
[tex] \sqrt{(x-3)} = y+4 [/tex]
[tex] y = \sqrt{(x-3)} - 4[/tex]
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) = [tex] [3, \infty)[/tex] (Option b)
Note: "[" or "]" bracket is used to INCLUDE the value. It means that 3 is included in the range.
