Question

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

Answer 1

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!