What is the Stationary point, x and y intercepts, point of inflextion of f(x)=(x-1)(x-2)^2 ?

f'(x) = x^3-5x^2+8x-4 =0
so stationary pint is 2 and 4/3 correct? What are the intercepts, points, and point of inflexion of the function and how would the graph look like?

1 Answer
Mar 30, 2018

Please see below.

Explanation:

.

#y=(x-1)(x-2)^2#

To find the stationary points we need to take the derivative of the function and set it equal to #0#.

We can either expand the second term and multiply the two parts together before taking the derivative or take it in the present form which is the product of two functions. As such, we use the product rule:

#dy/dx=(x-1)(2(x-2))+(x-2)^2=(x-2)(2x-2+x-2)#

#dy/dx=(x-2)(3x-4)=0#

#x-2=0, :. x=2#

#3x-4=0, :. x=4/3#

We can plug each one into the function to find the #y# coordinates:

#x=2, :. y=0#

#x=4/3, :. y=(4/3-1)(4/3-2)^2=1/3*4/9=4/27#

The stationary points are:

#(2,0)# is the minimum

#(4/3, 4/27)# is the maximum

We take the second derivative of the function and set it equal to #0# to find the inflection point:

#(d^2y)/dx^2=(x-2)(3)+(1)(3x-4)=3x-6+3x-4=6x-10#

#6x-10=0#

#x=5/3, :. y=(5/3-1)(5/3-2)^2=2/3*1/9=2/27#

#(5/3,2/27)# is the inflection point.

We set #x=0# to find the #y#-intercept;

#y=(-1)(-2)^2=-4#

#(0,-4)# is the #y#-intercept.

We set #y=0# to find the #x#-intercepts.

#(x-1)(x-2)^2=0#

#x=1# and

#x=2# two times.

#(1,0)# is an #x#-intercept.

The double root of #x=2# means the function touches the #x#-axis at that point but does not cross it.

The graph of the function is:

enter image source here