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 ).
Step-by-step explanation:
Circumference of a circle is 2piR
C=2piR and R=10
C=2(3.14)(10)
So the circumference is 62.8 cm
Answer:
A. { }
Step-by-step explanation:
Absolute value cannot be less than zero.
-4 < 0
There is no solution for x, ∈, R.
Hope this helps.
Answer:
-3z-11=11z^2
Step-by-step explanation: