Practice Problems

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Practice Problems
Compare the graph of the given quadratic function f with the graph of y = x2.
1) f(x) = (x - 2)2 + 3
First expand this function:
f(x) = (x - 2)(x – 2) + 3
= x2 – 4x – 4 + 3
= x2 – 4x – 1
If we compare these two function we see that the difference is in the (-4x – 1) term at the end. The x2 is
the same. So basically what does this end term do to the function?
The -4x - 1 term shifts the graph to the right and down, due to the negative signs before each term.
You can use the following site to compare: http://people.hofstra.edu/steven_r_costenoble/Graf/Graf.html
Determine if the function is even, odd, or neither.
2) f(x) = 2x5 + 2x3
To do this, you take the function and plug –x in for x, and then simplify. If you end up with the exact same
function that you started with (that is, if f(–x) = f(x)), then the function is even. If you end up with the exact
opposite of what you started with (that is, if f(–x) = –f(x)), then the function is odd.
Plug in –x for x
f(-x) = 2(-x)5 + 2(-x)3
= – 2x5 – 2x3
This is exactly opposite of what we started with so this function is odd!
Decide whether the relation defines a function.
3) {(-8, 2), (-8, 8), (-1, 8), (5, 6), (8, 7)}
The easiest way to determine this is to look at the first number in each pair. If any of them are the same,
then this is a relation, not a function. Looking above we see that there are two -8’s in this set (-8, 2) and (8, 8). Because this -8 domain is associated with two different ranges, we say this is a relation, not a
function.
5) y2 = 3x
???
Find the domain and range of the inverse of the given function.
7) f(x) = -4x - 2
First find the inverse of the function. Solve the equation for x:
y = -4x – 2
-4x = y + 2
x = - (y + 2) / 4
Now switch the variables x & y with each other.
y = - (x + 2) / 4
This is the inverse of the function.
The domain is the x values: all real numbers (±∞)
The range is the y values: all real numbers (±∞)
Compute and simplify the difference quotient
8) f(x) = 7x - 9
First find f(x + h):
f(x + h) = 7(x + h) – 9
Now plug into top equation:
Difference quotient = (f(x + h) – f(x)) / h
= (7(x + h) – 9) – (7x – 9) / h
= (7x +7h – 9 - 7x + 9) / h
= 7h / h
=7
simplify
Consider the function h as defined. Find functions f and g so that (f ∘ g)(x) = h(x).
9) h(x) =
(f ∘ g)(x) = f(g(x)) = h(x) = (8/x2) + 10
Possible solution for f(x) and g(x):
g(x) = x2
f(x) = (8/X) – 10
please see here for a more complete definition on fog functions:
http://www.helpalgebra.com/articles/compositionsoffunctions.htm
Find the requested value.
10) f(3) for f(x) = 3x + 1, if x < 1
f(3) undefined: (does not exist since x must be less the 1)
f(3) for f(x) = 3x, if 3 ≤ x ≤ 8
plug in and solve: f(3) = 3(3) = 9
f(3) for f(x) = 3 - 5x, if x > 8
f(3) undefined: (does not exist since x must be less the 1)
Determine whether the function is symmetric with respect to the y-axis, symmetric with respect to the xaxis, symmetric with respect to the origin, or none of these.
11) f(x) = -5x3 + 2x
To see if it is symmetric with respect to the y axis, plug in a value for +x and –x and see if you get the
same number.
Example: x = 1 & x = -1
f(1) = -5(1)3 + 2(1) = -3
f(-1) = -5(-1)3 + 2(-1) = + 3
Notice that we get two different answers (-3 and +3). So this is not symmetric about y axis. You can do
the same thing for x. Plug in negative and positive value for y, and then solve for x. If you do this you will
find we get a similar result as above, where the answers are the same but have opposite signs. Therefore
this function is not symmetric about the x axis.
When the answers are the same but have opposite signs, this means that it is not symmetric about x or y,
but it is symmetric about the origin. You can graph this function and see that is the case as well.
If the following defines a one-to-one function, find its inverse. If not, write "Not one-to-one."
12) {(-2, 4), (-1, 4), (0, 1), (1, -5)}
A function for which every element of the range of the function corresponds to exactly one element of the
domain is a one-to-one function. Look at the second number. If you have two that are the same range but
different domain, the function is not one-to-one. In this case we have two 4’s.
Function is not ONE-TO-ONE
If f is one-to-one, find an equation for its inverse.
13) f(x) = x3 - 5
It is one-to-one because every value of x you plug in gives you a different value for y.
The inverse is found the same way as above in problem 7. Solve for x, then switch the variables.
Perform the requested operation or operations.
14) f(x) =
, g(x) = 8x - 14
Find (f ∘ g)(x).
(f ∘ g)(x) = f(g(x)) = f(8x – 14) = Sqrt [(8x - 14) + 10] = Sqrt [8x – 4]
Graph the function.
16)
f(x) = 4x + 2 if x < -2
f(x) = x
if -2 ≤ x ≤ 3
f(x) = 3x-1
if x > 3
See below for graph:
y
6
4
2
X
6
4
2
2
4
6
2
4
6
Find the indicated composite for the pair of functions.
18) (g ∘ f)(x): f(x) =
, g(x) = 6x + 3
(g ∘ f)(x) = g(f(x))
= g((x – 3) / 6)
= 6((x – 3) / 6) + 3
=x–3+3
=x
Give the domain and range of the relation.
20) y = (x + 4)2 – 7
The domain (x’s): all real numbers will work with this relation
The range (y’s): If we examine this relation we can see that when x = -4 then the left term goes to zero
and we are left with -7
Y = (-4 + 4)2 – 7 = -7
This is the most negative this relation can get. We can not get a value less then -7. So the lower limit for
the range is -7. There is no upper limit on this relation.
So the range is: (-7, +∞)
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