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.
The above equation leads to
(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
Solve the above quadratic equation.
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.
More References and Links
Free Online Tutorials on Functions and Algebra
Questions on Functions with Solutions
Questions and Answers on Functions
