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

Apr 11, 2017

Step $1$: Determine the first derivative

$f ' \left(x\right) = 6 {x}^{2} - 24 x + 18$

Step $2$: Determine the critical numbers

These will occur when the derivative equals $0$.

$0 = 6 {x}^{2} - 24 x + 18$

$0 = 6 \left({x}^{2} - 4 x + 3\right)$

$0 = \left(x - 3\right) \left(x - 1\right)$

$x = 3 \mathmr{and} 1$

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

We select test points.

Test point 1: $x = 0$

$f ' \left(0\right) = 6 {\left(0\right)}^{2} - 24 \left(0\right) + 18 = 18$

Since this is positive, the function is uniformly increasing on $\left(- \infty , 1\right)$.

Test point 2: $x = 2$

$f ' \left(2\right) = 6 {\left(2\right)}^{2} - 24 \left(2\right) + 18 = - 6$

Since this is negative, the function is decreasing on $\left(1 , 3\right)$.

I won't select a test point for $\left(3 , \infty\right)$ 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.

$f ' ' \left(x\right) = 12 x - 24$

Step $5$: Determine the points of inflection

These will occur when $f ' ' \left(x\right) = 0$.

$0 = 12 x - 24$

$0 = 12 \left(x - 2\right)$

$x = 2$

Step $6$: Determine the intervals of concavity

Once again, we select test points.

Test point $1$: $x = 1$

$f ' ' \left(1\right) = 12 \left(1\right) - 24 = - 12$

This means that $f \left(x\right)$ concave down (Since it's negative) on$\left(- \infty , 2\right)$.

This also means that $f \left(x\right)$ is concave up on $\left(2 , \infty\right)$.

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

$f \left(x\right)$ 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

$f \left(0\right) = 2 {\left(0\right)}^{3} - 12 {\left(0\right)}^{2} + 18 x - 1 = - 1$

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!