A-Level Edexcel Maths Pure Graphs of Functions: The function f is defined by $$

The function f is defined by $$f : x ightarrow ext{ln}(4 - 2x), \, x < 2 \, \text{and} \, x \in \mathbb{R}.$$ a) Show that the inverse function of f is defined by $$f^{-1} : x ightarrow 2 - \frac{1}{2} e^x$$ and write down the domain of $f^{-1}$ - Edexcel - A-Level Maths Pure - Question 6 - 2007 - Paper 6

Question 6

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The function f is defined by $$f : x ightarrow ext{ln}(4 - 2x), \, x < 2 \, \text{and} \, x \in \mathbb{R}.$$ a) Show that the inverse function of f is defined b... show full transcript

Worked Solution & Example Answer:The function f is defined by $$f : x ightarrow ext{ln}(4 - 2x), \, x < 2 \, \text{and} \, x \in \mathbb{R}.$$ a) Show that the inverse function of f is defined by $$f^{-1} : x ightarrow 2 - \frac{1}{2} e^x$$ and write down the domain of $f^{-1}$ - Edexcel - A-Level Maths Pure - Question 6 - 2007 - Paper 6

Step 1

Show that the inverse function of f is defined by $f^{-1} : x \rightarrow 2 - \frac{1}{2} e^x$

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Answer

To find the inverse function, start with the function definition:

$y = \ln(4 - 2x)$

To solve for x in terms of y, exponentiate both sides: $e^y = 4 - 2x$

Rearranging gives: $2x = 4 - e^y$

Thus, $x = 2 - \frac{1}{2} e^y$

So the inverse is found to be:

$f^{-1}(x) : x \rightarrow 2 - \frac{1}{2} e^x$

The domain of $f^{-1}$ is all real numbers, or $(-\infty, \infty)$ , constrained by the original function's range, which will be $(\ln(4 - 2(2)), \infty)$ , or $(\ln(0), \infty)$ .

Step 2

Write down the range of $f^{-1}$

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Answer

The range of $f^{-1}$ is given by the domain of the original function $f$ , which is $(-\infty, 2)$ .

Step 3

Sketch the graph of $y = f^{-1}(x)$ and state the coordinates of the points of intersection with the x and y axes

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Answer

The graph of $y = f^{-1}(x)$ will be a downward curve starting from $(0, 2)$ . The intersections with the axes are:

At the x-axis, set $f^{-1}(x) = 0$ : $0 = 2 - \frac{1}{2} e^x \Rightarrow e^x = 2 \Rightarrow x = \ln(2)$ So the intersection with the x-axis is at $(\ln(2), 0)$ .
At the y-axis, set $x = 0$ : $f^{-1}(0) = 2 - \frac{1}{2} e^0 = 2 - \frac{1}{2} = 1$ The intersection with the y-axis is at $(0, 1)$ .

Step 4

Calculate the values of $x_1$ and $x_2$

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Answer

Using the iterative formula:

Start with $x_0 = -0.3$ : $x_1 = -\frac{1}{2} e^{-0.3} \approx -0.35092688$
Now use $x_1$ to find $x_2$ : $x_2 = -\frac{1}{2} e^{-0.35092688} \approx -0.3452$

Step 5

Find the value of k to 3 decimal places

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Answer

The value of k can be approximated as the limit of the iterative process. After several iterations: $k \approx -0.352$

Show that the inverse function of f is defined by $f^{-1} : x \rightarrow 2 - \frac{1}{2} e^x$

Write down the range of $f^{-1}$

Sketch the graph of $y = f^{-1}(x)$ and state the coordinates of the points of intersection with the x and y axes

Calculate the values of $x_1$ and $x_2$

Find the value of k to 3 decimal places

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