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f and g are functions such that $$f(x) = \frac{12}{\sqrt{x}}$$ and $$g(x) = 3(2x + 1)$$ (a) Find g(5) (b) Find g(f(9)) (c) Find g^{-1}(6) - Edexcel - GCSE Maths - Question 19 - 2020 - Paper 1

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f-and-g-are-functions-such-that--$$f(x)-=-\frac{12}{\sqrt{x}}$$-and-$$g(x)-=-3(2x-+-1)$$--(a)-Find-g(5)--(b)-Find-g(f(9))--(c)-Find-g^{-1}(6)-Edexcel-GCSE Maths-Question 19-2020-Paper 1.png

f and g are functions such that $$f(x) = \frac{12}{\sqrt{x}}$$ and $$g(x) = 3(2x + 1)$$ (a) Find g(5) (b) Find g(f(9)) (c) Find g^{-1}(6)

Worked Solution & Example Answer:f and g are functions such that $$f(x) = \frac{12}{\sqrt{x}}$$ and $$g(x) = 3(2x + 1)$$ (a) Find g(5) (b) Find g(f(9)) (c) Find g^{-1}(6) - Edexcel - GCSE Maths - Question 19 - 2020 - Paper 1

Step 1

Find g(5)

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Answer

To find g(5)g(5), we plug 55 into the function g(x)g(x):

g(5)=3(2(5)+1)=3(10+1)=3(11)=33g(5) = 3(2(5) + 1) = 3(10 + 1) = 3(11) = 33

Thus, g(5)=33g(5) = 33.

Step 2

Find g(f(9))

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Answer

First, calculate f(9)f(9):

f(9)=129=123=4f(9) = \frac{12}{\sqrt{9}} = \frac{12}{3} = 4

Now substitute 44 into g(x)g(x):

g(4)=3(2(4)+1)=3(8+1)=3(9)=27g(4) = 3(2(4) + 1) = 3(8 + 1) = 3(9) = 27

Therefore, g(f(9))=27g(f(9)) = 27.

Step 3

Find g^{-1}(6)

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Answer

To find the inverse g−1(6)g^{-1}(6), we start from the equation:

g(x)=6g(x) = 6

Substituting the expression for g(x)g(x) gives:

3(2x+1)=63(2x + 1) = 6

Dividing both sides by 33:

2x+1=22x + 1 = 2

Now, solving for xx:

2x=1x=122x = 1 \\ x = \frac{1}{2}

Hence, g−1(6)=12g^{-1}(6) = \frac{1}{2}.

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