# Plotting a function

1. Oct 8, 2008

### alpha01

1. The problem statement, all variables and given/known data

Using the local minima, local maxima and points of inflection of the following function, plot the graph:

f(x) = 9x4-11x3+3x2+1

3. The attempt at a solution

f(x) = 9x4-11x3+3x2+1

f ' (x) = 36x3-33x2+6x

= 3x(12x2-11x+2)

= 3x(3x-2)(4x-1)

therefore we have x = 0, 2/3, 1/4

now, finding the second derivative of the function we have:

f '' (x) = 108x2 - 66x + 6

plugging in our values of x into this, we have:

f '' (0) = 6

f '' (2/3) = 10

f '' (1/4) = - 15/4

therefore we have local minima x = 0, 2/3

and local maxima x = 1/4

I have also found 1 point of inflection to be: 0.029...

I have plotted the function using an online function plotter, and the local minima/maxima dont appear to correspond with the actual graph.

Could someone please show in which step i went wrong?

2. Oct 8, 2008

### gabbagabbahey

You've calculated the correct local max/mins so if you plot from x=-1/3 to x=4/3 you should be able to see them. If you plot a larger range of x-values, these local max/mins will get blurred out.

Your inflection point on the other hand is incorrect. How did you arrive at it?

3. Oct 8, 2008

### alpha01

Thanks.. The point of inflection is 9.. i worked it out again.

4. Oct 8, 2008

### gabbagabbahey

I think there are two points of inflection, and neither is at x=9....maybe you should show me what you are doing to get your answer, so I can see where you are going wrong.

5. Oct 8, 2008

### alpha01

0 = 108x2-66x+6

0 = 108x2-54x-12x+6

0 = 54x(2x-1)-6(2x-1)

6(2x-1) = 54x(2x-1)

6 = (54x(2x-1)) / 2x-1

6 = 54x

x = 1/9

sorry i meant 1/9 not 9

is this still incorrect? as u said .. there should be 2 of them..

6. Oct 8, 2008

### Mentallic

From $$0 = 54x(2x-1)-6(2x-1)$$ do you realize this can be factored? From the previous line you should've noticed the quadratic.
The reason why you are missing one of the points of inflection (whatever that is, I never learnt it) is because you divided by one of the solutions.

When you divided by $$2x-1$$ this means that $$x\neq \frac{1}{2}$$ since you cannot divide by 0, but 0.5 is one of the solutions. You just need to be sure you approach the quadratic correctly so as to find all the solutions.