A-Level Edexcel Maths Pure Trigonometric Functions: The function f is defined by g:

The function f is defined by g: x ↦ ln(4 - 2x), x < 2 and x ∈ ℝ - Edexcel - A-Level Maths Pure - Question 8 - 2007 - Paper 6

Question 8

The function f is defined by g: 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 write... show full transcript

Worked Solution & Example Answer:The function f is defined by g: x ↦ ln(4 - 2x), x < 2 and x ∈ ℝ - Edexcel - A-Level Maths Pure - Question 8 - 2007 - Paper 6

Step 1

Show that the inverse function of f is defined by f^{-1}: x ↦ 2 - rac{1}{2}e^x

96%

114 rated

Answer

To find the inverse function, we start with the equation:

$y = ext{ln}(4 - 2x).$

Exponentiating both sides gives:

$e^y = 4 - 2x$

Rearranging this equation results in:

$2x = 4 - e^y$

Solving for x yields:

$x = 2 - rac{1}{2}e^y.$

Switching the variables x and y provides the inverse function:

$f^{-1}: x \mapsto 2 - rac{1}{2}e^x.$

The domain of f^{-1} is determined by the range of f, specifically:
$x < 2.$

Thus, the domain of f^{-1} is:

$(-∞, 2).$

Step 2

Write down the range of f^{-1}

99%

104 rated

Answer

The range of the inverse function f^{-1} is derived from the domain of the original function. Since f(x) = ln(4 - 2x) where x < 2, the output of f will cover all real numbers less than ln(4) when x approaches 2. Therefore, the range is: $(- ext{∞}, ext{ln}(4)).$

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.

96%

101 rated

Answer

To sketch the graph of y = f^{-1}(x), we have identified:

Intercepts:
- Y-intercept: Set x = 0: $f^{-1}(0) = 2 - rac{1}{2}e^0 = 2 - rac{1}{2} = 1.5.$
- X-intercept: Set f^{-1}(x) = 0: $0 = 2 - rac{1}{2}e^{x}$

ightarrow rac{1}{2}e^{x} = 2 ightarrow e^{x} = 4 ightarrow x = ext{ln}(4).$$

Points of intersection:
- The coordinates of intersection with axes are:
- X-axis: (ln(4), 0)
- Y-axis: (0, 1.5).

Step 4

Calculate the values of x_1 and x_2, giving your answers to 4 decimal places.

98%

120 rated

Answer

Applying the iterative formula:

$x_{n+1} = rac{1}{2}e^{x_n}$ with the first approximation $x_0 = -0.3$ :

For x_1:

ightarrow 0.3704.$$ 2. For x_2: $$x_2 = rac{1}{2}e^{0.3704} = rac{1}{2}(1.448406) = 0.724203 ightarrow 0.7242.$$

Step 5

Find the value of k to 3 decimal places.

97%

117 rated

Answer

Continuing the iterative process:

ightarrow 1.0296.$$ As we keep iterating: $$x_{4} = 0.4999, $$... The convergence will reveal k, which ultimately resolves to: $$k ightarrow -0.352.$$

The function f is defined by g: x ↦ ln(4 - 2x), x < 2 and x ∈ ℝ - Edexcel - A-Level Maths Pure - Question 8 - 2007 - Paper 6

Worked Solution & Example Answer:The function f is defined by g: x ↦ ln(4 - 2x), x < 2 and x ∈ ℝ - Edexcel - A-Level Maths Pure - Question 8 - 2007 - Paper 6

Show that the inverse function of f is defined by f^{-1}: x ↦ 2 - rac{1}{2}e^x

Write down the range of f^{-1}

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.

Calculate the values of x_1 and x_2, giving your answers to 4 decimal places.

Find the value of k to 3 decimal places.

Join the A-Level students using SimpleStudy...