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:
Hello
Step-by-step explanation:
The domain is limited with 2 lines parallel: -1 ≤ x ≤ 1
and the disk ? (inside of a circle) of center (0,0) and radius 2

The answer is actually D. The limit on the left side is computing the derivative at x = a. The right side value of 7 tells us that f ' (a) = 7
Answer:
x = 3.87
Step-by-step explanation:
Using the right angle altitude theorem, we know that all three triangles are congruent, so the lengths of corresponding sides of the triangles are in proportion..
AD/DB = DB/DC
15/x = x/1
x² = 15
x = √15
x = 3.87
Answer:
The equation for the axis of symmetry is x=1.5.
Step-by-step explanation:
It is given that a quadratic function passes through the points ( − 5 , − 8 ) and ( 8 , − 8 ) .
In the given points y-coordinates are same, i.e., -8. It means both the points lie on the horizontal line y=-8.
If a quadratic function passes through two points (a,c) and (b,c), then the equation for the axis of symmetry is

According to the given points a=-5, b=8 and c=-8. Put these value in the above formula.



Therefore the equation for the axis of symmetry is x=1.5.