How do you find all local maximum and minimum points using the second derivative test given #y=x^3-9x^2+24x#?

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Answer 1 · 2017-03-16T14:39:45

We have a local maximum at #(2,20)#
We have a local minimum at #(4,16)#

Let's calculate the first derivative

#y=x^3-9x^2+24x#

Let #f(x)=x^3-9x^2+24x#

#f'(x)=dy/dx=3x^2-18x+24#

#=3(x^2-6x+8)=3(x-2)(x-4)#

The critical points are when #dy/dx=0#

That is, when #x=2# and #x=4#

Now, we calculate the second derivative

f''(x)=(d^2y)/dx^2=6x-18#

When #f''(x)=(d^2y)/dx^2=0#, we have an inflexion point at #x=3#

Now,

We calculate

#f''(2)=12-18=-6#

As, #f''(2) <0#, we have a local maximum at #(2,20)#

#f''(4)=24-18=6#

As, #f''(2) >0#, we have a local minimum at #(4,16)#

graph{x^3-9x^2+24x [-36.67, 45.55, -6.95, 34.14]}