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The function f is defined by $$ f(x) = 4 + 3^{-x}, \, x \in \mathbb{R} $$ 10 (a) Using set notation, state the range of f 10 (b) (i) Using set notation, state the domain of f^{-1} 10 (b) (ii) Find an expression for f^{-1}(x) 10 (c) The function g is defined by $$ g(x) = 5 - \sqrt{x}, \, (x \in \mathbb{R}: x > 0) $$ 10 (c) (i) Find an expression for gf(x) 10 (c) (ii) Solve the equation gf(x) = 2, giving your answer in an exact form. - AQA - A-Level Maths: Pure - Question 10 - 2017 - Paper 1

Question 10

$The-function-f-is-defined-by--$$-f(x)-=-4-+-3^{-x},-\,-x-\in-\mathbb{R}-$$--10-(a)-Using-set-notation,-state-the-range-of-f--10-(b)-(i)-Using-set-notation,-state-the-domain-of-f^{-1}--10-(b)-(ii)-Find-an-expression-for-f^{-1}(x)--10-(c)-The-function-g-is-defined-by--$$-g(x)-=-5---\sqrt{x},-\,-(x-\in-\mathbb{R}:-x->-0)-$$--10-(c)-(i)-Find-an-expression-for-gf(x)--10-(c)-(ii)-Solve-the-equation-gf(x)-=-2,-giving-your-answer-in-an-exact-form.-AQA-A-Level Maths: Pure-Question 10-2017-Paper 1.png$

The function f is defined by $$ f(x) = 4 + 3^{-x}, \, x \in \mathbb{R} $$ 10 (a) Using set notation, state the range of f 10 (b) (i) Using set notation, state the... show full transcript

Worked Solution & Example Answer:The function f is defined by $$ f(x) = 4 + 3^{-x}, \, x \in \mathbb{R} $$ 10 (a) Using set notation, state the range of f 10 (b) (i) Using set notation, state the domain of f^{-1} 10 (b) (ii) Find an expression for f^{-1}(x) 10 (c) The function g is defined by $$ g(x) = 5 - \sqrt{x}, \, (x \in \mathbb{R}: x > 0) $$ 10 (c) (i) Find an expression for gf(x) 10 (c) (ii) Solve the equation gf(x) = 2, giving your answer in an exact form. - AQA - A-Level Maths: Pure - Question 10 - 2017 - Paper 1

Step 1

Using set notation, state the range of f

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Answer

To determine the range of the function $f(x) = 4 + 3^{-x}$ , we observe the behavior of the function as $x$ tends towards positive and negative infinity. As $x \to -\infty$ , $3^{-x} \to \infty$ , hence $f(x) \to \infty$ . As $x \to \infty$ , $3^{-x} \to 0$ , therefore $f(x) \to 4$ . The lowest value of the function is 4 and it extends to infinity. Thus, the range of $f$ is:

$ext{Range of } f = \{ y \in \mathbb{R} : y > 4 \}$

Step 2

Using set notation, state the domain of f^{-1}

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Answer

The domain of the inverse function $f^{-1}$ corresponds to the range of the original function $f$ . From part (a), we found the range of $f$ to be $\{ y \in \mathbb{R} : y > 4 \}$ . Therefore, the domain of $f^{-1}$ is:

$\text{Domain of } f^{-1} = \{ x \in \mathbb{R} : x > 4 \}$

Step 3

Find an expression for f^{-1}(x)

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Answer

To find the inverse function $f^{-1}(x)$ , we start by interchanging $x$ and $y$ . We have:

$x = 4 + 3^{-y}$

Rearranging this gives:

$x - 4 = 3^{-y}$

Taking the logarithm on both sides:

$-y \log(3) = \log(x - 4)$

Thus, we have:

$y = -\frac{\log(x - 4)}{\log(3)}$

Consequently, the expression for the inverse function is:

$f^{-1}(x) = -\frac{\log(x - 4)}{\log(3)}$

Step 4

Find an expression for gf(x)

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Answer

To find the expression for $gf(x)$ , we substitute $f(x)$ into $g(x)$ :

$g(f(x)) = g(4 + 3^{-x}) = 5 - \sqrt{4 + 3^{-x}}$

Step 5

Solve the equation gf(x) = 2, giving your answer in an exact form

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Answer

To solve for $gf(x) = 2$ :

$5 - \sqrt{4 + 3^{-x}} = 2$

Rearranging gives:

$\sqrt{4 + 3^{-x}} = 3$

Squaring both sides:

$4 + 3^{-x} = 9$

This simplifies to:

$3^{-x} = 5$

Taking logarithms:

$-x \log(3) = \log(5)$

Thus:

$x = -\frac{\log(5)}{\log(3)}$

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