Find the coordinates of the point of inflection?

Question

1. A curve has equation y = x(x - 4)^2

Find the coordinates of the maximum point and the point of inflection.

2. Find the coordinates of the point on the graph of y = x^2 - x at which the tangent is parallel to the line y = 5x.

Step by step explanations please!

Answer 1

1) y = x^3 - 8x^2 +16x so y' = 3x^2 - 16x and y" = 6x - 16 =0 at the reqd point so x = 16/6 or 8/3. Use this value in the given eqn and find y-coordinate.

2) y' = 2x-1 = 5 so x = 3 and so y =6 so (3, 6) is the reqd pt.

Answer 2

1. differentiate once (using the multiplication rule)

y' = 1*(x-4)^2+x*2*(x-4)= x^2-8x+16+2x^2-4x

compare to zero to find critical points

y' =3x^2-12x+16=0

x= (12+-sqrt(144-4*3*16))/2*3

no real solutions - therefore there isn't any maximum nor inflection point on the real plane.

2. differentiate and compare to 5

y'=2x-1=5

2x=6

x=3

y(3)=3^2-3=9-3=6

therefore the point is (3,6)

Answer 3

1.

y = x(x - 4)^2

y' = (x - 4 )^2 + 2x(x - 4) = (x - 4)(x - 4 + 2x) = (x - 4)(3x - 4)

y" = 3(x - 4) + (3x - 4) = 6x - 16

The x-coordinate of the maximum point satisfies both conditions y' = 0 and y" < 0

(x -4)(3x -4) = 0 and 6x - 16 < 0 ==> x = 4/3

The x-coordinate of the inflexion satisfies y" = 0

6x - 16 = 0 ==> x = 8/3

Maximum point: (4/3, ...)

Inflexion:(8/3,....)

2.

y = x^2 - x

y' = 2x - 1

y' = 5 when 2x - 1 = 5, x = 3, then y = 6

The answer is (3, 6)