Math Problem Statement

the inverse of f(x) = 2^x / 1+2^x is 1. In (y/y-1) - In (2) 2. In base 2 (y/y-1) 3. none of the above 4. In(2) / 1+In(2) 5. In(y) + In(y+1) 6. cannot be determined

Solution

Let's find the inverse of the function f(x)=2x1+2xf(x) = \frac{2^x}{1 + 2^x} and analyze the provided options.

Step 1: Express y=f(x)y = f(x)

y=2x1+2xy = \frac{2^x}{1 + 2^x}

Step 2: Solve for xx in terms of yy

First, multiply both sides by 1+2x1 + 2^x to eliminate the denominator: y(1+2x)=2xy(1 + 2^x) = 2^x y+y2x=2xy + y \cdot 2^x = 2^x

Now, isolate the term with 2x2^x on one side: y=2xy2xy = 2^x - y \cdot 2^x y=2x(1y)y = 2^x(1 - y)

Divide both sides by 1y1 - y: y1y=2x\frac{y}{1 - y} = 2^x

Now, take the natural logarithm (ln) of both sides to solve for xx: x=ln(y1y)/ln(2)x = \ln\left(\frac{y}{1 - y}\right) / \ln(2)

So, the inverse function f1(y)f^{-1}(y) is: f1(y)=ln(y1y)ln(2)f^{-1}(y) = \frac{\ln\left(\frac{y}{1 - y}\right)}{\ln(2)}

Step 3: Compare with the provided options

  1. ln(yy1)ln(2)\ln\left(\frac{y}{y - 1}\right) - \ln(2):

    • This expression is incorrect because ln(yy1)\ln\left(\frac{y}{y - 1}\right) is not the correct form; it should be ln(y1y)\ln\left(\frac{y}{1 - y}\right).
  2. lnbase 2(yy1)\ln_{\text{base 2}}\left(\frac{y}{y - 1}\right):

    • This is also incorrect for the same reason as above. The argument inside the logarithm should be y1y\frac{y}{1 - y}, not yy1\frac{y}{y - 1}.
  3. None of the above:

    • This could be correct if none of the options match the correct inverse.
  4. ln(2)1+ln(2)\frac{\ln(2)}{1 + \ln(2)}:

    • This expression doesn't relate to the inverse function we found.
  5. ln(y)+ln(y+1)\ln(y) + \ln(y + 1):

    • This is also incorrect because it doesn’t match the form of the inverse function.
  6. Cannot be determined:

    • The inverse can be determined, so this is not correct.

Correct Answer:

  • The correct answer is 3. None of the above since none of the provided options accurately represent the inverse function f1(y)=ln(y1y)ln(2)f^{-1}(y) = \frac{\ln\left(\frac{y}{1 - y}\right)}{\ln(2)}.

Would you like more details or have any questions? Here are some related questions:

  1. How do you derive the inverse of a function?
  2. Why is ln(y1y)\ln\left(\frac{y}{1 - y}\right) used in finding the inverse?
  3. How do logarithms help in solving for inverses of exponential functions?
  4. What is the difference between natural logarithm and logarithm with other bases?
  5. Can the inverse of a function always be expressed in terms of elementary functions?
  6. How would the inverse change if the base of the exponential function were different?
  7. What are common pitfalls when solving for the inverse of a function?
  8. How can you verify if a function and its inverse are correct?

Tip: When finding the inverse of a function, carefully isolate the variable of interest and ensure all steps logically follow. Checking your final expression by composing the original and inverse functions can confirm correctness.

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Math Problem Analysis

Mathematical Concepts

Exponential Functions
Logarithms
Inverse Functions

Formulas

Inverse function formula

Theorems

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Suitable Grade Level

Advanced High School