How do you sketch the graph #f(x)=2x^3-12x^2+18x-1#?

1 Answer
Apr 11, 2017

Step #1#: Determine the first derivative

#f'(x) = 6x^2 - 24x + 18#

Step #2#: Determine the critical numbers

These will occur when the derivative equals #0#.

#0 = 6x^2 - 24x + 18#

#0 = 6(x^2 - 4x + 3)#

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

#x = 3 or 1#

Step #3#: Determine the intervals of increase/decrease

We select test points.

Test point 1: #x = 0#

#f'(0) = 6(0)^2 - 24(0) + 18 = 18#

Since this is positive, the function is uniformly increasing on #(-oo, 1)#.

Test point 2: #x = 2#

#f'(2) = 6(2)^2 - 24(2) + 18 = -6#

Since this is negative, the function is decreasing on #(1, 3)#.

I won't select a test point for #(3, oo)# because I know the function is increasing on the interval. The point #x = 3# is referred to as a turning point because it goes from decreasing to increasing or vice versa.

Step #4#: Determine the second derivative

This is the derivative of the first derivative.

#f''(x) = 12x - 24#

Step #5#: Determine the points of inflection

These will occur when #f''(x) = 0#.

#0 = 12x- 24#

#0 = 12(x - 2)#

#x = 2#

Step #6#: Determine the intervals of concavity

Once again, we select test points.

Test point #1#: #x = 1#

#f''(1) = 12(1) - 24 = -12#

This means that #f(x)# concave down (Since it's negative) on#(-oo, 2)#.

This also means that #f(x)# is concave up on #(2, oo)#.

Step #7#: Determine the x/y- intercept

#f(x)# doesn't have any rational factors, so we'll forget about the x-intercepts (as a result, you would need Newton's Method or a similar method to find the x-intercepts).

The y-intercept is

#f(0) = 2(0)^3 - 12(0)^2 + 18x - 1 = -1#

If you can't connect the graph, you could always make a table of values. In the end, you should get a graph similar to the following. graph{2x^3- 12x^2 + 18x - 1 [-10, 10, -5, 5]}

Hopefully this helps!