Evaluation and composition of function
• The letter f in the function notation f(x) as the name of the function. Instead of
using the equation to describe the function,
• we can write . Here, f is the name of the function and f(x) is the
value of the function at x. So is the value of the
function at 2.
• The notation affords a convenient way of prompting the evaluation of a function
for a particular value of x.
• Any letter can be used as the independent variable in a function. So the above
function could be written .
• The independent variable in a function is just a place holder. The function could
be written without a variable as follows
• The function can be viewed as an input/output operation.
• In addition to plugging numbers into functions, we can plug expressions into
functions. Plugging y + 1 into the function yields
• Plug other expressions in terms of x into a function. Plugging 2x into the function
• The variable x in the function is being replaced by the same variable. But the x in
function is just a placeholder. If the placeholder were removed from the function,
the substitution would appear more natural. In , we plug 2x into
the left side f(2x) and it returns the right side
• We have plugged numbers into functions and expressions into functions; now
let’s plug in other functions. Since a function is identified with its expression.
• In the example above with and 2x, let’s call 2x by the name g(x).
In other words, g(x) = 2x. Then the composition of f with g (that is plugging g into
• The notation f(g(x)) on the test. But you probably will see one or more problems
that ask you perform the substitution. For another example, let and
let . Then
• The composition of functions merely substitutes one function into another, these
problems can become routine. Notice that the composition operation f(g(x)) is
performed from the inner parentheses out, not from left to right. In the operation
f(g(2)), the number 2 is first plugged into the function g and then that result is
plugged in the function f.
• A function can also be composed with itself. That is, substituted into itself. Let
Given f(x) = 2x + 3 and g(x) = –x2 + 5, find ( f o g)(x).
A) –2x2 + 13
B) 2x2 + 13
C) –2x2 – 13
D) 2x2 – 13
plugging the formula for g(x) into the formula for f(x)
( f o g)(x) = f (g(x))
= f (–x2 + 5)
= 2( )+3 ... setting up to insert the input formula
= 2(–x2 + 5) + 3
= –2x2 + 10 + 3
= –2x2 + 13
Given f(x) = 2x + 3 and g(x) = –x2 + 5, find ( f o g)(1)
( f o g)(1) = f (g(1))
= f (–( )2 + 5) ... setting up to insert the original input
= f (–(1)2 + 5)
= f (–1 + 5)
= f (4)
= 2( ) + 3 ... setting up to insert the new input
= 2(4) + 3
1) The graph of y = f(x) is shown to the right. If f(-1) = v, then which one of the
following could be the value of f(v) ?
2) In the function above, if f(k) = 2, then which one of the following could be a value
of k ?
3) Given f(x) = 2x + 3 and g(x) = –x2 + 5, find (g o f )(x).
(A) 4x2 + 12x – 4
(B) –4x2 – 12x – 4
(C) –4x2 – 12x + 4
(D) 4x2 + 12x + 4
(E) –4x2 + 12x + 4
4) Given f(x) = 2x + 3 and g(x) = –x2 + 5, find ( f o f )(x)
(A) 4x – 9
(B) 4x + 7
(C) 4x + 9
(D) -4x + 9
(E) -4x - 9
5) Given f(x) = 2x + 3 and g(x) = –x2 + 5, find (g o f )(1)
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