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

Question

Scarlett Allison · Accepted Answer

#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]}

Evelyn Agard · Answer

undefined To graph $ f(x) = x^3 - 3x^2 - 9x + 6 $ using the information from the first derivative, follow these steps:

1. Find the critical points by setting the first derivative equal to zero and solving for $ x $.
2. Determine the intervals where the first derivative is positive or negative to identify increasing and decreasing sections.
3. Locate any local maxima or minima by analyzing the behavior of the first derivative around the critical points.
4. Use the behavior of the first derivative to sketch the graph of the function.

To summarize:
- Find critical points by solving $ f'(x) = 0 $.
- Determine intervals of increase and decrease by analyzing the sign of $ f'(x) $.
- Determine local extrema by analyzing the behavior of $ f'(x) $ around critical points.
- Sketch the graph using this information.