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 $g(f(x))$. 10 (c) (ii) Solve the equation $g(f(x)) = 2$, giving your answer in an exact form.

A-Level AQA Maths Pure Functions: The function f is defined by $f

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 - AQA - A-Level Maths Pure - Question 10 - 2017 - Paper 1

Question 10

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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 dom... 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 - 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 first recognize that $3^x$ takes all positive real values since the base is greater than 1. Therefore, as $x$ varies, $3^x$ goes from 0 to $+ ext{∞}$ . Thus, the minimum value of $f(x)$ occurs when $x o - ext{∞}$ , approaching 4. Consequently, the range of $f$ is:

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

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 function $f$ . Since the range of $f$ is all values greater than 4, the domain of $f^{-1}$ is:

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

Step 3

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

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Answer

To find $f^{-1}(x)$ , we start with the equation:

$y = 4 + 3^x$

Rearranging gives:

$3^x = y - 4$

Next, we can take the logarithm of both sides:

$x = \log_3(y - 4)$

Thus, we have the inverse function:

$f^{-1}(x) = \log_3(x - 4)$

Step 4

Find an expression for $g(f(x))$.

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Answer

To find $g(f(x))$ , we substitute $f(x)$ into the function $g$ :

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

Step 5

Solve the equation $g(f(x)) = 2$, giving your answer in an exact form.

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Answer

To solve the equation:

$g(f(x)) = 2$

we start with:

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

Rearranging gives:

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

Next, we square both sides:

$4 + 3^x = 9$

Then, isolate $3^x$ :

$3^x = 9 - 4 = 5$

Taking logarithms:

$x = \log_3(5)$

Therefore, the solution is:

$x = \log_3(5)$

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 - AQA - A-Level Maths Pure - Question 10 - 2017 - Paper 1

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 - AQA - A-Level Maths Pure - Question 10 - 2017 - Paper 1

Using set notation, state the range of f.

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

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

Find an expression for $g(f(x))$.

Solve the equation $g(f(x)) = 2$, giving your answer in an exact form.

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