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 = (fg)(x) = ?
The expression (fg)(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.
