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:
He would have to wait for 5 weeks
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$5 x $5 = $25
$25 + $10.50 = $35.50
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d. (5,-2)
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On the original segment (black endpoints), the (red) point is the most accurate to being 2/3 the distance from endpoint (-3,8)
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I dont understand your question rither
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sorry I would help is I neew
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The sentence you need is ##### ###### ### ### #####
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