How do you graph #f(x)=x^3-3x^2-9x+6# using the information given by the first derivative?

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Answer 1 · 2016-11-02T10:37:28

#f(x)# has a local maximun at #(-1, 11)# and a local minimum at #(3, -21)#

#f(x) = x^3-3x^2-9x+6#

#f'(x) = 3x^2-6x-9#

#f(x)# will have turning points where #f'(x) =0#

#f(x)=0 -> 3x^2-6x-9 = 0#

#x^2-2x-3=0#

#(x-3)(x+1) = 0#

#:. f(x)# has turning points at #x = 3# and #x=-1#

Now consider, #f''(x) = 6x-6#

#f''(3) = 12 > 0 -> x=3# is a local minimum
#f''(-1) = -12 < 0 -> x=-1# is a local maximum

These points can be seen on the graph of #f(x)# below:

graph{x^3-3x^2-9x+6 [-43.74, 46.25, -24.88, 20.13]}