GCSE Edexcel Maths Geometry Toolkit: The functions g and h are such t

The functions g and h are such that g(x) = √(2x - 5) h(x) = 1/x (a) Find g(16) - Edexcel - GCSE Maths - Question 20 - 2022 - Paper 2

Question 20

The functions g and h are such that g(x) = √(2x - 5) h(x) = 1/x (a) Find g(16). (b) Find hg'(x) Give your answer in terms of x in its simplest form.

Worked Solution & Example Answer:The functions g and h are such that g(x) = √(2x - 5) h(x) = 1/x (a) Find g(16) - Edexcel - GCSE Maths - Question 20 - 2022 - Paper 2

Step 1

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Answer

To find g(16), we substitute x = 16 into the function g:

$g(16) = ext{√(2(16) - 5)}$
Calculating the expression inside the square root:

$2(16) - 5 = 32 - 5 = 27$
Thus,

$g(16) = ext{√27} = 3√3$

Step 2

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Answer

To find hg'(x), we first need to calculate the derivative of h(x).

Given h(x) = 1/x, we find h'(x):

$h'(x) = - rac{1}{x^2}$

Next, we compute hg'(x) as follows:

$hg'(x) = g(h(x)) imes h'(x)$

Substituting h(x) into g:

$g(h(x)) = gigg( rac{1}{x}igg) = ext{\sqrt(2( rac{1}{x}) - 5)}$

Now, substituting into hg'(x):

$hg'(x) = gigg( rac{1}{x}igg) imes h'(x) = ext{√igg( rac{2}{x} - 5igg)} imes - rac{1}{x^2}$

Thus, in its simplest form, we express hg'(x) as:

$hg'(x) = - rac{ ext{√igg( rac{2}{x} - 5igg)}}{x^2}$