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

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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+3xf(x) = 4 + 3^x, we first recognize that 3x3^x takes all positive real values since the base is greater than 1. Therefore, as xx varies, 3x3^x goes from 0 to +ext+ ext{∞}. Thus, the minimum value of f(x)f(x) occurs when xoextx o - ext{∞}, approaching 4. Consequently, the range of ff is:

extRangeoff={y:y>4,  yR} 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 f1f^{-1} corresponds to the range of the function ff. Since the range of ff is all values greater than 4, the domain of f1f^{-1} is:

Domain of f1={x:x>4,  xR}\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 f1(x)f^{-1}(x), we start with the equation:

y=4+3xy = 4 + 3^x

Rearranging gives:

3x=y43^x = y - 4

Next, we can take the logarithm of both sides:

x=log3(y4)x = \log_3(y - 4)

Thus, we have the inverse function:

f1(x)=log3(x4)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))g(f(x)), we substitute f(x)f(x) into the function gg:

g(f(x))=g(4+3x)=54+3xg(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))=2g(f(x)) = 2

we start with:

54+3x=25 - \sqrt{4 + 3^x} = 2

Rearranging gives:

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

Next, we square both sides:

4+3x=94 + 3^x = 9

Then, isolate 3x3^x:

3x=94=53^x = 9 - 4 = 5

Taking logarithms:

x=log3(5)x = \log_3(5)

Therefore, the solution is:

x=log3(5)x = \log_3(5)

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