Math Problem Statement

Please answer the following questions

Given the graph of a function f(x) below, answer the following questions: (a) The domain of f(x) is (b) The range of f(x) is (c) f(1) = (d) f(-3) = (e) (f ∘ f)(-1) = (f) Find a nonzero value of x for which g(x) = 1/(1 + f(x)) is not defined.

Solution

Let's analyze the given graph and answer each question in detail based on the function f(x)f(x).

(a) The domain of f(x)f(x)

The domain of a function represents all the possible values of xx for which f(x)f(x) is defined. From the graph:

  • The function exists from x=6x = -6 to x=1x = 1 with some open circles indicating points where the function is not defined at specific xx-values.
  • The function is undefined at x=3x = -3 and x=0x = 0 (indicated by open circles at these points).

Thus, the domain of f(x)f(x) is:

Domain of f(x)=(6,3)(3,0)(0,1]\text{Domain of } f(x) = (-6, -3) \cup (-3, 0) \cup (0, 1]

(b) The range of f(x)f(x)

The range is the set of all possible yy-values the function takes. Based on the graph:

  • The minimum yy-value is 3-3 (reached at x=1x = -1).
  • The maximum yy-value seems to be unbounded as the graph increases for x>1x > 1.
  • The function skips some values, like it doesn’t exist at y=1y = -1 (corresponding to x=0x = 0).

Thus, the range of f(x)f(x) is:

Range of f(x)=[3,2)(2,)\text{Range of } f(x) = [-3, 2) \cup (2, \infty)

(c) f(1)f(1)

From the graph, at x=1x = 1, the function has a solid point at y=4y = 4.

f(1)=4f(1) = 4

(d) f(3)f(-3)

At x=3x = -3, there is an open circle at y=0y = 0, indicating that the function is not defined at x=3x = -3.

f(3) is undefinedf(-3) \text{ is undefined}

(e) (ff)(1)(f \circ f)(-1)

This represents f(f(1))f(f(-1)). First, we need to find f(1)f(-1):

  • From the graph, f(1)=2f(-1) = -2.

Next, we evaluate f(2)f(-2):

  • From the graph, f(2)=0f(-2) = 0.

Thus:

(ff)(1)=f(f(1))=f(2)=0(f \circ f)(-1) = f(f(-1)) = f(-2) = 0

(f) Find a nonzero value of xx for which g(x)=11+f(x)g(x) = \frac{1}{1 + f(x)} is not defined.

The expression g(x)g(x) will be undefined when the denominator is zero, i.e., when:

1+f(x)=0f(x)=11 + f(x) = 0 \quad \Rightarrow \quad f(x) = -1

From the graph, f(x)=1f(x) = -1 at x=0x = 0.

Thus, the nonzero value of xx is:

x=0(but since we need a nonzero value, there are none that fit this condition)x = 0 \quad \text{(but since we need a nonzero value, there are none that fit this condition)}

However, strictly speaking, x=0x = 0 is the solution if non-zero wasn't specified.


Let me know if you want further details or have any questions! Here are 5 related questions for deeper understanding:

  1. How would the graph change if f(x)f(x) had a discontinuity at more points?
  2. What would the domain be if f(x)f(x) was defined at x=0x = 0?
  3. How do we calculate (ff)(x)(f \circ f)(x) in general for composite functions?
  4. How does the range of a function influence the solutions of g(x)g(x)?
  5. How can we interpret the undefined points in terms of limits?

Tip: Always check for open and closed points in graphs to identify where a function is defined or undefined.

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Math Problem Analysis

Mathematical Concepts

Functions
Domain and Range
Composite Functions
Undefined Expressions

Formulas

g(x) = 1/(1 + f(x))
(f ∘ f)(x) = f(f(x))

Theorems

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Suitable Grade Level

Grades 9-12