# Questions on Functions with Solutions

A set of questions on functions and their properties are presented. Detailed solutions to the questions are also provided.

## Questions

Question 1: f is a function defined by

.
Find f(0), f(2) and f(4).

Question 2: Is y a function of x in the following equation

-x 2 + y 2 - 2 = 0?

Question 3: Find the domain of function f defined by

f(x) = √ ( x 2 + 1 ) + 2

Question 4: f is a function defined by

f(x) = |x - 1| + 3.

Find f(-9).

Question 5: f and g are functions defined by

f(x) = 2x + 2 and g(x) = √ (x + 3).

Find (f o g)(6).

Question 6: Find all real values of x such that f(x) = 0 given that f is a function defined by

f(x) = (x 2 + 2 x - 3) / (x - 1)

Question 7: Find all real values of x such that f(x) = g(x) where f and g are functions given by

f(x) = 3x + √(x) and g(x) = 2x + 6

Question 8:
is y as a function of x in the following equation?

|y + 1| = x 2 + 2

Question 9: Find the domain of function f defined by

f(x) = 1 / √ (|x - 1|)

### Solutions to the Above Questions

Solution to Question 1:
f(0) is found by substituting x by 0 in the formula
f(x) = 2x + 4 since 0 is less than 2.

f(0) = 4
f(2) is found by substituting x by 2 in the formula
f(x) = 2x + 4 since 2 is less than or equal to 2.

f(2) = 2*2 + 4 = 8
f(4) is found by substituting x by 4 in the formula
f(x) = 2x - 1 since 4 is greater than 2.

f(4) = 2*4 - 1 = 7

Solution to Question 2:
To answer the question, we need to solve the given equation for y.
y = + or - √ (x 2 +2)
y is not a function of x because for one value of the independent variable x (input) we obtain two values for the dependent variable y (output).

Solution to Question 3:
The expression inside the square root is positive for any real value of x; therefore the domain of function f is the set of all real numbers.

Solution to Question 4:
Evaluate f(-9) by substituting x by -9 in the formula of the function.
f(-9) = |-9 - 1| + 3 = 10 + 3 = 13

Solution to Question 5:
(f o g)(6) is calculated as follows
(f o g)(6) = f(g(6))
Let us first evaluate g(6)
g(6) = √ (6 + 3) = 3
We now substitute g(6) by 3 in f(g(6))
f(g(6)) = f(3)
Finally evaluate f(3)
f(3) = 2×3 + 2 = 8
and (f o g)(6) = 8.

Solution to Question 6:
f(x) = 0 gives the following equation
(x 2 + 2 x - 3) / (x - 1) = 0
The denominator is not equal to 0 for values of x not equal to 1.
(x 2 + 2 x - 3) = 0
Solve the above equation by factoring
(x - 1)(x + 3) = 0
The above equation has two solutions
x = 1 and x = -3.
x = 1 is not a solution to f(x) = 0 because It is a value of x that makes the denominator of f(x) equal to zero. So the only value of x for which f(x) = 0 is -3.

Solution to Question 7:
g(x) = g(x) leads to an equation.
3x + √(x) = 2x + 6
Rewrite the equation as follows
√(x) = - x + 6
Square both sides and simplify
x = (- x + 6) 2
x = x 2 + 36 - 12 x
Rewrite in standard form.
x 2 - 13 x + 36 = 0
x = 4 and x = 9
Since we squared both sides of the equation, extraneous solutions may be introduced but can be eliminated by checking. After checking, x = 4 is the only value of x that makes f(x) = g(x).

Solution to Question 8:
Solve the given equation for y.
y = -1 - (x 2 + 2) or y = -1 + (x 2 + 2)
The two solutions mean that for one value of x we obtain two values of y and y is not a function of x.

Solution to Question 9:
The domain of function f is the set of all real values of x for which f(x) is real. |x - 1| is always positive except when x = 1 which makes it equal to 0. But |x - 1| is in the denominator. Therefore the domain of f is the set of all real numbers except 0.