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The function f is defined by f : x ↦ ln(4 - 2x), x < 2 and x ∈ R - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 6
Question 7
The function f is defined by f : x ↦ ln(4 - 2x), x < 2 and x ∈ R. (a) Show that the inverse function of f is defined by f^{-1} : x ↦ 2 - rac{1}{2} e^{x} and wri... show full transcript
Worked Solution & Example Answer:The function f is defined by f : x ↦ ln(4 - 2x), x < 2 and x ∈ R - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 6
Step 1
Show that the inverse function of f is defined by f^{-1} : x ↦ 2 - rac{1}{2} e^{x}
Answer
To find the inverse function, we start with the function: Exponentiating both sides gives: Rearranging yields: So, we have: x = 2 - rac{1}{2} e^y This indicates the inverse function is: f^{-1} : x ↦ 2 - rac{1}{2} e^{x}.
Next, we find the domain of f^{-1}. The domain of f was x < 2, so the range of f will be: Therefore, the domain of f^{-1} is:
Step 2
Step 3
In the space provided on page 16, sketch the graph of y = f^{-1}(x). State the coordinates of the points of intersection with the x and y axes.
Answer
To sketch the graph of y = f^{-1}(x), it is essential to note the following:
- The graph should show a decreasing curve.
- It intersects the x-axis when: y = 0 ext{ or } 2 - rac{1}{2} e^x = 0, which leads to: The intersection point is (ln(4), 0).
- For the y-intercept, set x = 0: f^{-1}(0) = 2 - rac{1}{2} e^0 = 2 - rac{1}{2} = 1. Thus, the y-intercept is (0, 1).
Step 4
Calculate the values of x_1 and x_2, giving your answers to 4 decimal places.
Answer
Starting with the iterative formula: x_{n+1} = - rac{1}{2} e^{x_n}, x_0 = -0.3 For x_1: x_1 = - rac{1}{2} e^{-0.3} ≈ -0.35092688 Thus x_1 to 4 decimal places is -0.3509.
Now calculating x_2: x_2 = - rac{1}{2} e^{-0.35092688} ≈ -0.35201767 Thus x_2 to 4 decimal places is -0.3520.
Step 5