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Pure Maths 12 today........

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Does anyone know how to answer Q10(b) the minimum value of the curve and also to prove that its a minimum?

The question was about the minimum value of the 'Gradient' not the curve,So DY/DX of the curve is the equation of the tangent,D^2Y/DX^2 is the second deff of curve but first deff of Gradient,so you get D^2Y/DX^2 and equal it to 0,get the X substitute it into DX/DY to get minimum value of 'Gradient',and to make sure that it's a minimum value I got Third deff which was second deff of gradient got a positive product which means it's minimum.
 
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domain was equal or more than 1 or only more than 1
because i wrote equal or more
 
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The range was not given,I Had to get it then place it as the domain of inverse,domain of function was x>3 not the inverse ? can't really remember if you still remember the function we can solve it again :D
 
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The range was not given,I Had to get it then place it as the domain of inverse,domain of function was x>3 not the inverse ? can't really remember if you still remember the function we can solve it again :D
yeah uare right igot the same...
 
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Show that in the 2nd question was kind of easy,still took me about 10 mins to get it
 
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I made so many stupid mistakes!! T___T
for the area of the playgroud i think that they asked to write the minimum value of the area but i instead left it at x=15 T___T
I was too hasty and did not entirely read the question
 
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