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A function, $f$, is given by $f(x) = ext{ } oot{3}{x} + 8$ - Scottish Highers Maths - Question 8 - 2019

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Question 8

A-function,-$f$,-is-given-by-$f(x)-=--ext{-}-oot{3}{x}-+-8$-Scottish Highers Maths-Question 8-2019.png

A function, $f$, is given by $f(x) = ext{ } oot{3}{x} + 8$. The domain of $f$ is $1 ext{ ≤ } x ext{ ≤ } 1000, x ext{ ∈ } ext{R}$. The inverse function, $f^{-1}$... show full transcript

Worked Solution & Example Answer:A function, $f$, is given by $f(x) = ext{ } oot{3}{x} + 8$ - Scottish Highers Maths - Question 8 - 2019

Step 1

Find $f^{-1}(x)$

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Answer

To find the inverse function, start by setting the function equal to xx:

oot{3}{y} + 8$$ 1. Rearrange the equation to isolate the cubic root:

oot{3}{y} = x - 8$$

  1. Cube both sides to eliminate the cubic root: y=(x−8)3y = (x - 8)^3

Thus, the inverse function can be expressed as: f−1(x)=(x−8)3f^{-1}(x) = (x - 8)^3

Step 2

State the domain of $f^{-1}$

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Answer

The domain of the inverse function f−1f^{-1} is determined by the range of the original function ff. Since the original function f(x)=oot3x+8f(x) = oot{3}{x} + 8 has a domain of xext∈[1,1000]x ext{ ∈ } [1, 1000], we evaluate:

  • At x=1x = 1, f(1)=oot31+8=9f(1) = oot{3}{1} + 8 = 9.
  • At x=1000x = 1000, f(1000)=oot31000+8=10+8=18f(1000) = oot{3}{1000} + 8 = 10 + 8 = 18.

Therefore, the range of ff is [9,18][9, 18], which means the domain of the inverse function f−1f^{-1} is: [9,18][9, 18].

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