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Need help in p1 O/N/06 q:10 (iii) (iv) (v) and M/J/07 q:11 (ii) (iv)
Q11)
ii) replace f ' (x) with y : y = 6 (2x + 3)
2x+3 = 6/y
2x= 6/y - 3
x = (6/y - 3 ) / 2
replace x with f ' x and y with x ; so f ' (x) = (6/x - 3) / 2
Now to find the domain you need to visualize the graph of f ' (x). You can infer that x has a small range. Now, where does the graph of f ' (x) cut the x axis?
0 = (6/x - 3) /2
x= 2
and where does the graph meet y axis? You can assume that the graph meets y axis.
y = (6/x - 3x/x) / 2 (I have taken the LCM of the terms in the bracket)
y = (6/0 - 0/0) Note: Anything divided by zero is infinity. So the graph starts at (0 , infinity) and ends at ( 2, 0)
0<x<(or equal to) 2
Note that the x is greater than 0 , not greater than or equal to. It is because for any value of x greater than 0, the graph exists but on x=0 , y becomes infinity. I hope its kind of clear