# Sketching a graph using derivatives

• Oct 30th 2012, 07:04 PM
DylanRenke
Sketching a graph using derivatives
sketch f(x)=(x2-1)3

Not sure where i am going wrong but when i look at the actual graph of this i cannot figure out the concavity. Could someone please help me out.

so far i have found f'(x)=6x5-12x3+6x and f''(x)=30x4-36x2+6

Thank you

Dylan Renke
• Oct 30th 2012, 07:48 PM
MarkFL
Re: Sketching a graph using derivatives
We are given:

$f(x)=(x^2-1)^3$

If I were going to sketch the graph of the given function, I would first note that the function is even, which means it is reflected across the y-axis.

It has a positive root at (1,0), and so must have one at (-1,0). It has a y-intercept at (0,-1). The roots are of odd multiplicity, so they will pass through the x-axis, rather than just being tangent to it.

There are no asymptotes.

Now, differentiating, we find:

$f'(x)=3(x^2-1)^2(2x)=6x(x^2-1)^2$

The root $x=0$ is of odd multiplicity, so we can expect an extremum there, but the other two roots are of even multiplicity, so they will not be at extrema.

We then find that the function is decreasing on the intervals $(-\infty,-1)\,\cup\,(-1,0)$ and increasing on $(0,1)\,\cup\,(1,\infty)$.

By the first derivative test, we can state then that there is a global minimum at (0,-1).

Differentiating again, we find:

$f''(x)=6x(2(x^2-1)(2x))+6(x^2-1)^2=6(x^2-1)(5x^2-1)$

Here, all of the roots of of odd multiplicity so we can expect all of the roots to be at points of inflection.

The critical numbers are $x=-1,-\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}},1$

Testing the intervals made by these numbers, we find:

$(-\infty,-1)$ concave up.

$\left(-1,-\frac{1}{\sqrt{5}} \right)$ concave down.

$\left(-\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}} \right)$ concave up.

$\left(\frac{1}{\sqrt{5}},1 \right)$ concave down.

$(1,\infty)$ concave up.

We now have enough information to make a reasonable sketch of the graph.
• Oct 31st 2012, 06:39 AM
DylanRenke
Re: Sketching a graph using derivatives
can you please explain how you came up with the concavity because when i did it i did not come up with the same answers as you did.
• Oct 31st 2012, 07:14 AM
HallsofIvy
Re: Sketching a graph using derivatives
You had, originally, f''= 30x^4- 36x^2+ 6 and to find points of inflection, where the convexity might change, set that equal to 0:
30x^4- 36x^2+ 6= 0. An obvious first step is to divide that by 6: 5x^4- 6x^2+ 1= 0. You should be able to see that this can be factored as (5x^2- 1)(x^2- 1)= 0 so that 5x^2- 1= 0 or x^2- 1= 0 as MarkFL2 said. If that is not immediatley clear, note that only the even powers of x are in this equation and let y= x^2 so the equation is 5y^2- 6y+ 1= 0. and solve that using the quadratic formula or completing the square to get y= 1 and y= 1/5 so that x= 1, -1, 1/sqrt(5), -1/sqrt(5).

Then you can choose a number in each of the intervals to see whether the second derivative is positive or negative. Alternatively, you can use that information to write f''= 30(x+ 1)(x+ 1/sqrt(5))(x- 1/sqrt(5))(x- 1). If x< -1 then each factor us x minus a larger number and so all negative. There are four factors so the product is positive. Is -1< x< -1/sqrt(5) so we now have the first factor positive, the other three negative- the product is now positive. You should be able to see that the sign of f'' alternates as we "pass" each of those zeros.