Math Problem Statement

The graph of an exponential function g(x) = a(b)^x + c is shown. Use the graph to answer the following questions: A. What is the value of c? Give a reason for your answer. B. Consider the following and explain how you know: i. Is a > 0 or a < 0? ii. Is b > 1 or 0 < b < 1? iii. Use the points shown on the graph to find the values for a and b. C. The inverse function of g is a logarithmic function. Find g⁻¹(x). D. Identify the domain and range of the inverse function from part C.

Solution

Let's analyze the uploaded question and work through the problems step by step:


Problem A:

  • The function given is g(x)=a(b)x+cg(x) = a(b)^x + c.
  • To determine the value of cc, note that cc represents the horizontal asymptote of the graph.
  • Observing the graph, the asymptote is at y=4y = -4, so: c=4c = -4

Reason: The function levels off as x±x \to \pm\infty, and the horizontal asymptote is evident in the graph.


Problem B:

Part i: a>0a > 0 or a<0a < 0?

  • Observing the graph, the curve decreases as xx increases (it has a downward trend).
  • The decreasing behavior indicates that a>0a > 0 and 0<b<10 < b < 1.

Part ii: b>1b > 1 or 0<b<10 < b < 1?

  • Since the graph is decreasing (exponential decay), 0<b<10 < b < 1.

Part iii: Determine aa and bb using points from the graph:

  1. Use the points (0,2.5)(0, 2.5) and (1,2)(1, -2) to solve for aa and bb.

    From g(0)=2.5g(0) = 2.5: g(0)=a(b)0+c    2.5=a(1)4    a=6.5g(0) = a(b)^0 + c \implies 2.5 = a(1) - 4 \implies a = 6.5

    From g(1)=2g(1) = -2: g(1)=a(b)1+c    2=6.5(b)4    6.5b=2    b=26.50.3077g(1) = a(b)^1 + c \implies -2 = 6.5(b) - 4 \implies 6.5b = 2 \implies b = \frac{2}{6.5} \approx 0.3077

Thus, a=6.5a = 6.5 and b0.3077b \approx 0.3077.


Problem C:

The inverse of g(x)=6.5(0.3077)x4g(x) = 6.5(0.3077)^x - 4 is a logarithmic function:

  1. Start with y=6.5(0.3077)x4y = 6.5(0.3077)^x - 4.
  2. Solve for xx: y+4=6.5(0.3077)xy + 4 = 6.5(0.3077)^x y+46.5=(0.3077)x\frac{y + 4}{6.5} = (0.3077)^x x=log0.3077(y+46.5)x = \log_{0.3077}\left(\frac{y + 4}{6.5}\right)

The inverse function is: g1(x)=log0.3077(x+46.5)g^{-1}(x) = \log_{0.3077}\left(\frac{x + 4}{6.5}\right)


Problem D:

Domain and Range of the Inverse Function:

  • Domain: The original range of g(x)g(x), which is (4,)(-4, \infty).
  • Range: The original domain of g(x)g(x), which is (,)(-\infty, \infty).

Final Summary:

  1. c=4c = -4.
  2. a=6.5a = 6.5, b0.3077b \approx 0.3077, 0<b<10 < b < 1, a>0a > 0.
  3. g1(x)=log0.3077(x+46.5)g^{-1}(x) = \log_{0.3077}\left(\frac{x + 4}{6.5}\right).
  4. Domain of g1(x)g^{-1}(x): (4,)(-4, \infty).
  5. Range of g1(x)g^{-1}(x): (,)(-\infty, \infty).

Would you like further clarification on any part or details about logarithmic functions? Here are some follow-up questions to explore:

  1. Why does cc represent the horizontal asymptote?
  2. How does bb affect whether the exponential graph increases or decreases?
  3. What happens to the inverse function if b>1b > 1?
  4. Can you explain the steps to find the domain and range of g1(x)g^{-1}(x) in more detail?
  5. How do transformations affect the graph of exponential and logarithmic functions?

Tip: To confirm your work with logarithmic bases, you can rewrite the function as ln(x)/ln(b)\ln(x)/\ln(b) if your calculator lacks logarithms with arbitrary bases.

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Math Problem Analysis

Mathematical Concepts

Exponential Functions
Logarithmic Functions
Function Inverses
Domain and Range

Formulas

g(x) = a(b)^x + c
Inverse: g⁻¹(x) = log_b((x + c) / a)

Theorems

Properties of exponential and logarithmic functions
Inverse function rule

Suitable Grade Level

Grades 10-12