MHB F(x) = x2-7 and g(x) = x- 3, find (f º g )(x) [2]

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To find (f º g)(x), substitute g(x) into f(x), resulting in (x - 3)² - 7. For (g º f)(x), substitute f(x) into g(x), yielding x² - 7 - 3. The discussion also touches on finding the inverse functions f⁻¹(x) and g⁻¹(x), but the focus remains on the composition of the functions. The user expresses confusion about the notation and seeks clarification on how to proceed with the calculations. Understanding function composition is crucial for solving these problems effectively.
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Let f(x) = x2-7 and g(x) = x- 3
Find:
i. (f º g )(x) [2]
ii. (g º f) (x) [2]
iii. f-1 (x) = g-1(x)
 
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trevor said:
Let f(x) = x2-7 and g(x) = x- 3
Find:
i. (f º g )(x) [2]
ii. (g º f) (x) [2]
iii. f-1 (x) = g-1(x)

Hi trevor. What have you tried so far? :)
 
Joppy said:
Hi trevor. What have you tried so far? :)

I am clueless
 
(f o g)(x) means f(g(x)). That means, you replace the value in f(x) with g(x), thus your x2 - 7 will become (g(x))2 - 7. Can you continue from here?
 

Thread 'Trigonometry problem of interest'

Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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