Study Notes: Function Image Problems - Techniques

Study Notes: Function Image Problems -
Techniques

Institution: California State University, Northridge（Northridge, CA）

Course: Math 250（Vector Calculus）

Instructor: David Klein

Instructor Time: Last Tuesday



This section focuses on techniques for solving function image problems, specifically
using parity (odd/even properties), special values, and limits.
The context introduces three key techniques for solving problems related to
function graphs:
1. Parity : Determining if a function is odd, even, or neither.

o Odd functions have graphs symmetric about the origin. Their domain
must be symmetric about the origin, and �( − �) =− �(�).

o Even functions have graphs symmetric about the y-axis. Their
domain must be symmetric about the origin, and �( − �) = �(�).

2. Special Value Method : Substituting specific, convenient values for the
variable (often 0, 1, -1, or values that simplify the expression) to test against
the graphs properties or to eliminate options.

3. Limit Method : Examining the behavior of the function as the variable
approaches certain values, especially infinity or zero, to understand
asymptotes or the functions trend.



Lets break down the provided examples:

Technique 01: Determining Function Graph from its Formula
Example 1-1:

, ln|�|
 Problem: The approximate graph of the function �(�) = �
on the interval
( − ∞, 0) ∪ (0, + ∞) is given.

 Analysis using Parity:

o The domain is � ≠ 0, which is symmetric about the origin.
o Lets check for odd/even properties:
ln|−�| ln|�| ln|�|
�( − �) = −�
= −�
=− �
=− �(�).

o Since �( − �) =− �(�), the function is odd. This means its graph is
symmetric about the origin.
o Elimination: Options B and D are eliminated because their graphs do
not exhibit origin symmetry.

 Analysis using Special Value Method:

o Lets pick a value, say � = �.
ln|�| ln� 1
o �(�) = �
= �
= �.

o Looking at the remaining options (A and C), option A shows a positive
value at � = �, while option C shows a negative value.
1
o Conclusion: Since �(�) = � > 0, option A is selected.

 Analysis using Limit Method:
ln�
o As � → 0+ , ln� →− ∞ and � → 0+ . The ratio �
tends to −∞.
ln|�|
o As � → 0− , ln|�| →− ∞ and � → 0− . The ratio �
tends to +∞.

o This behavior (approaching +∞ as � → 0− ) is consistent with option A.

 Answer: A

Example 1-2:
 Problem: The graph of the function �(�) = �sin� is given.

 Analysis using Parity:

o The domain is ℝ, which is symmetric about the origin.
o �( − �) = ( − �)sin( − �) = ( − �)( − sin�) = �sin� = �(�).

, o Since �( − �) = �(�), the function is even. Its graph is symmetric
about the y-axis.
o Elimination: Option A is eliminated because its graph is not
symmetric about the y-axis.

 Analysis using Special Value Method:

o Let � = �. �(�) = �sin(�) = � ⋅ 0 = 0.
� � � � � �
o Let � = 2. �( 2 ) = 2 sin( 2 ) = 2 ⋅ 1 = 2.
� � � � � �
o Let � =− 2. �( − 2 ) =− 2 sin( − 2 ) =− 2 ⋅ ( − 1) = 2.

o Consider option C: At � = �, the graph shows a value close to 0, but it
seems to be negative. This contradicts �(�) = 0. So, C is eliminated.
�
o Consider option D: At � = 2, the graph shows a positive value. At � =−
� � �
2
, it also shows a positive value. This is consistent with �( 2 ) = 2.

o Conclusion: Option D fits the special values.

 Analysis using Limit Method:

o As � → ∞, the function oscillates between −� and �.

 Analysis using Monotonicity (from context):

o The context mentions that for � ∈ (0, �), the function is monotonically
increasing. Lets check this.

o �′(�) = sin� + �cos�.
�
o For � ∈ (0, 2 ), sin� > 0 and cos� > 0, so �′ (�) > 0.
�
o For � ∈ ( 2 , �), sin� > 0 and cos� < 0. The sign of �′(�) depends on the
values.

o The context states that for � ∈ (0, �), the function is monotonically
increasing. This eliminates option B, which shows a decrease in part
of this interval.

 Answer: D

Practice Problem 1:
 Problem: The approximate graph of the function �(�) = �cos� is given.

 Analysis using Parity: