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The provided solution covers questions about the domain of a logarithmic function, finding its inverse, transformations, and solving inequalities using reflections. Which original question does this s

Mathematics

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Answers

3 parts
5.1

x>0\mathbf{x > 0}

5.2

y=3−x\mathbf{y = 3^{-x}}

5.4

x≥27\mathbf{x \ge 27}

Step-by-step solution

10 steps

Part 5.1

Write down the domain of ff.

The function is f(x)=−log⁡3xf(x) = -\log_3 x. For a logarithmic function log⁡bx\log_b x, the argument xx must be greater than 0.
The domain of ff is x>0\boxed{\mathbf{x > 0}}.

Part 5.2

y=3−x\mathbf{y = 3^{-x}}

Write down the equation of f−1f^{-1} in the form y=...y=....

  1. Replace f(x)f(x) with yy.

    y=−log⁡3xy = -\log_3 x
  2. Swap xx and yy to find the inverse.

    x=−log⁡3yx = -\log_3 y
  3. Isolate the logarithm.

    −x=log⁡3y-x = \log_3 y
  4. Convert the logarithmic equation to an exponential equation.

    y=3−xy = 3^{-x}

    The equation of f−1f^{-1} is y=3−x\boxed{\mathbf{y = 3^{-x}}}.

Part 5.3· 2 steps

Describe the transformation from f−1f^{-1} to hh if h(x)=3x−5h(x)=3^{x}-5.

  1. Identify f−1(x)f^{-1}(x).

    From 5.2, f−1(x)=3−xf^{-1}(x) = 3^{-x}.

  2. Compare f−1(x)f^{-1}(x) with h(x)=3x−5h(x) = 3^x - 5.

    To change 3−x3^{-x} to 3x3^x, we replace xx with −x-x. This is a reflection in the y-axis.
    To change 3x3^x to 3x−53^x - 5, we subtract 5 from the function. This is a vertical shift downwards by 5 units.
    The transformations are:

    • Reflection in the y-axis.
    • Vertical shift 5 units downwards.

Part 5.4

x≥27\mathbf{x \ge 27}

Use the graph of ff to solve for xx if k(x)k(x) is the reflection of ff about the x-axis and k(x)≥3k(x) \ge 3.

  1. Determine the equation of k(x)k(x).

    The function f(x)=−log⁡3xf(x) = -\log_3 x.
    A reflection about the x-axis means multiplying the function by −1-1.

    k(x)=−f(x)=−(−log⁡3x)=log⁡3xk(x) = -f(x) = -(-\log_3 x) = \log_3 x
  2. Set up the inequality k(x)≥3k(x) \ge 3.

    log⁡3x≥3\log_3 x \ge 3
  3. Solve the inequality for xx.

    Convert the logarithmic inequality to an exponential inequality. Since the base 3>13 > 1, the inequality direction remains the same.

    x≥33x \ge 3^3 x≥27x \ge 27
  4. Consider the domain of k(x)k(x).

    For k(x)=log⁡3xk(x) = \log_3 x, the domain requires x>0x > 0.
    Combining x≥27x \ge 27 with x>0x > 0, the solution is x≥27x \ge 27.
    The values of xx are x≥27\boxed{\mathbf{x \ge 27}}.

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