Let A( t , f( t ) ) be the point(s) at which the graph of the function has a horizontal tangent => f ' ( t ) = 0.
But, f ' ( x ) = [ ( x^2 ) ' * ( x - 1 ) - ( x^2 ) * ( x - 1 )' ] / ( x - 1 )^2 =>
f ' ( x ) = [ 2x( x - 1 ) - ( x^2 ) * 1 ] / ( x - 1 )^2 => f ' ( x ) = ( x^2 - 2x ) / ( x - 1 )^2;
f ' ( t ) = 0 <=> t^2 - 2t = 0 <=> t * ( t - 2 ) = 0 <=> t = 0 or t = 2 => f ( 0 ) = 0; f ( 2 ) = 4 => A 1 ( 0 , 0 ) and A 2 ( 2 , 4 ).
Answer: The answer would be 15
Step-by-step explanation:
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A.
So for 1 hour you pay 10 dolars then for 3 you pay 30 then for 11 you pay 110 then for 20 you pay 200.
For B and C I dont Know sorry.
6 and 10 because 6 times 10 is 60 and 6 plus 6 plus 20 plus 10 is 32
Answer:
he should chose compounded annually because he would have and extra 32.12 dollars at the end of the 4 years
Step-by-step explanation:
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