Math Problem Statement

solve

The functions f and g are defined by: f(x) = e^(2x), g(x) = x + 1. (a) Calculate f^(-1)(3) × g^(-1)(3). (b) Show that (f ∘ g)^(-1)(3) = ln(√3) - 1.

Solution

Let's break down the problem step by step:

Given:

  • f(x)=e2xf(x) = e^{2x}
  • g(x)=x+1g(x) = x + 1

Part (a): Calculate f1(3)×g1(3)f^{-1}(3) \times g^{-1}(3).

1. Find f1(x)f^{-1}(x):

To find the inverse of f(x)=e2xf(x) = e^{2x}, we express y=f(x)=e2xy = f(x) = e^{2x}, and solve for xx in terms of yy.

y=e2xln(y)=2xx=12ln(y)y = e^{2x} \quad \Rightarrow \quad \ln(y) = 2x \quad \Rightarrow \quad x = \frac{1}{2} \ln(y)

Thus, f1(x)=12ln(x)f^{-1}(x) = \frac{1}{2} \ln(x).

2. Find f1(3)f^{-1}(3):

Using the inverse function: f1(3)=12ln(3)f^{-1}(3) = \frac{1}{2} \ln(3)

3. Find g1(x)g^{-1}(x):

For g(x)=x+1g(x) = x + 1, the inverse is straightforward. Set y=g(x)=x+1y = g(x) = x + 1, and solve for xx:

x=y1x = y - 1

Thus, g1(x)=x1g^{-1}(x) = x - 1.

4. Find g1(3)g^{-1}(3):

g1(3)=31=2g^{-1}(3) = 3 - 1 = 2

5. Multiply f1(3)×g1(3)f^{-1}(3) \times g^{-1}(3):

Now, multiply the two values: f1(3)×g1(3)=12ln(3)×2=ln(3)f^{-1}(3) \times g^{-1}(3) = \frac{1}{2} \ln(3) \times 2 = \ln(3)

So, the answer to part (a) is ln(3)\ln(3).


Part (b): Show that (fg)1(3)=ln(3)1(f \circ g)^{-1}(3) = \ln(\sqrt{3}) - 1.

First, we need to calculate the composition fgf \circ g, find its inverse, and then evaluate it at 3.

1. Find fgf \circ g:

The composition (fg)(x)(f \circ g)(x) means f(g(x))f(g(x)), so: f(g(x))=f(x+1)=e2(x+1)=e2x+2=e2e2xf(g(x)) = f(x + 1) = e^{2(x+1)} = e^{2x+2} = e^2 \cdot e^{2x}

Thus, fg(x)=e2e2xf \circ g(x) = e^2 \cdot e^{2x}.

2. Find the inverse of fg(x)f \circ g(x):

Set y=fg(x)=e2e2xy = f \circ g(x) = e^2 \cdot e^{2x}, and solve for xx:

y=e2e2xye2=e2xln(ye2)=2xy = e^2 \cdot e^{2x} \quad \Rightarrow \quad \frac{y}{e^2} = e^{2x} \quad \Rightarrow \quad \ln\left(\frac{y}{e^2}\right) = 2x

ln(y)2=2xx=12ln(y)1\ln(y) - 2 = 2x \quad \Rightarrow \quad x = \frac{1}{2} \ln(y) - 1

Thus, (fg)1(x)=12ln(x)1(f \circ g)^{-1}(x) = \frac{1}{2} \ln(x) - 1.

3. Evaluate (fg)1(3)(f \circ g)^{-1}(3):

Now substitute x=3x = 3 into the inverse function: (fg)1(3)=12ln(3)1(f \circ g)^{-1}(3) = \frac{1}{2} \ln(3) - 1

Notice that 12ln(3)\frac{1}{2} \ln(3) is the same as ln(3)\ln(\sqrt{3}), so:

(fg)1(3)=ln(3)1(f \circ g)^{-1}(3) = \ln(\sqrt{3}) - 1

This proves the desired result.


Final Answers:

  • Part (a): ln(3)\ln(3)
  • Part (b): ln(3)1\ln(\sqrt{3}) - 1

Would you like further details or clarifications on any part of the solution?

Here are 5 related questions:

  1. How do you generally find the inverse of an exponential function?
  2. What are the steps for composing two functions?
  3. Can you explain why f1(x)×g1(x)f^{-1}(x) \times g^{-1}(x) worked out this way?
  4. How would you find the inverse of a more complex composite function?
  5. What is the significance of natural logarithms in inverse calculations?

Tip: Always check your composition when dealing with inverse functions to ensure that the result makes sense when plugging back into the original!

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Math Problem Analysis

Mathematical Concepts

Inverse functions
Exponential functions
Function composition

Formulas

Inverse of f(x) = e^(2x) is f^(-1)(x) = (1/2)ln(x)
Inverse of g(x) = x + 1 is g^(-1)(x) = x - 1
(f ∘ g)^(-1)(x) = (1/2)ln(x) - 1

Theorems

Properties of inverse functions
Logarithmic identities

Suitable Grade Level

Grades 11-12 (Advanced High School Mathematics or Pre-Calculus)