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:
26
Step-by-step explanation:
Devaughn = 2x
Sydney = x
x + 2x = 78
3x = 78
x = 78 / 3
x = 26
Hope that helps!
1) A
3) B
4) A
5)A
sorry that i dont have all the answers, hope this helped
D(-1,-1), E(-8,-4), F(-8,-8)
Answer:
1.50x=18
x=12
Step-by-step explanation:
You have to divide 1.50 on both sides and 1.50x/1.50= x and 18/1.50=12 so x=12
