Concavity, Inflection Points, and Intervals of Increase and Decrease

[ardentmed]

Hey guys,

Need some more help again. I'll keep it brief.

This thread is only for question one. Please ignore 2ab.

I started off by computing f'(x) via the product rule and calculating the zeros, at which point I got:
x= 1/√ 5 and x=√(3)

However, I'm at a loss as to what I should do next. Taking the derivative of f'(x) =4x(10x^2 - 8) , which means that the inflection points are found at x=0, and x=√ 4/√ 5

Am I on the right track, or did I make a clerical error up to this point?


What I don't fully grasp is whether I should be confirming my answers with the derivative test or if stating them is sufficient.
Thanks in advance.

 

[MarkFL]

Okay, we are given:

$$f(x)=\left(x^2-1\right)^3$$

To find the intervals of increase/decrease, we will need to analyze the first derivative. To differentiate $f$, we will need to use the power rule along with the chain rule:

$$f^{\prime}(x)=3\left(x^2-1\right)^2(2x)=6x\left(x^2-1\right)^2$$

Now, we can easily see that the roots of $f^{\prime}$ are:

$$x\in\{-1,0,1\}$$

Now, if we observe that the non-zero roots are of multiplicity 2, then we know the sign of $f^{\prime}$ will not change across these roots. And since the root $x=0$ is of multiplicity 1, we know the sign will change across this root.

So, we need only text one sub-interval, and so let's pick a value for $x$ from $\left(-\infty,-1\right)$, say $-2$ and see what the sign of $f^{\prime}$ is:

$$f^{\prime}(-2)=6(-2)\left((-2)^2-1\right)^2<0$$

Thus we may then conclude:

$\left(-\infty,-1\right)$: $f$ is decreasing.

$\left(-1,0\right)$: $f$ is decreasing.

$\left(0,1\right)$: $f$ is increasing.

$\left(1,\infty\right)$: $f$ is increasing.

Now, to find the intervals of concave up and concave down, we need to perform a similar analysis on the second derivative. This time you will want to use the product rule...give it a try and I will be glad to look over your work. :D

 

[ardentmed]

Okay, we are given:

$$f(x)=\left(x^2-1\right)^3$$

To find the intervals of increase/decrease, we will need to analyze the first derivative. To differentiate $f$, we will need to use the power rule along with the chain rule:

$$f^{\prime}(x)=3\left(x^2-1\right)^2(2x)=6x\left(x^2-1\right)^2$$

Now, we can easily see that the roots of $f^{\prime}$ are:

$$x\in\{-1,0,1\}$$

Now, if we observe that the non-zero roots are of multiplicity 2, then we know the sign of $f^{\prime}$ will not change across these roots. And since the root $x=0$ is of multiplicity 1, we know the sign will change across this root.

So, we need only text one sub-interval, and so let's pick a value for $x$ from $\left(-\infty,-1\right)$, say $-2$ and see what the sign of $f^{\prime}$ is:

$$f^{\prime}(-2)=6(-2)\left((-2)^2-1\right)^2<0$$

Thus we may then conclude:

$\left(-\infty,-1\right)$: $f$ is decreasing.

$\left(-1,0\right)$: $f$ is decreasing.

$\left(0,1\right)$: $f$ is increasing.

$\left(1,\infty\right)$: $f$ is increasing.

Now, to find the intervals of concave up and concave down, we need to perform a similar analysis on the second derivative. This time you will want to use the product rule...give it a try and I will be glad to look over your work. :D

Alright. This took me quite a while, but I'm pretty sure I computed the correct answer(s).

I performed the second derivative test for:

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

And computed:

x$\in$ {-1, -1/√5 ,1/√5 ,1}

As the roots. Therefore, I performed the second derivative test with arbitrary points between the roots. If it matters, I used -2, -1/2, 1/10, 1/2, and 2 as the x values to test with f''.

Furthermore, I concluded that f is concave up from:

$$(-\infty,-1)u(-1/√5,1/√5)u(1,\infty) $$

Moreover, f is concave down from: $$(-1,-1/√5)u(1/√5,1) $$


As for inflection points, I computed:

f(-1)=0
f(1)=0
f(1/√5)= -64/125
f(-1/√5)= -64/125
f(1)=0

Am I on the right track? Thank you so much for typing out that well-thought-out response.

 

[MarkFL]

Yes, you computed the second derivative correctly, correctly identified the intervals of concave up/down, and found the points of inflection.

Observing that the second derivative had only roots of multiplicity 1 would have allowed you to test only 1 interval, and then let the sign of the second derivative alternate across the intervals. Also, if you notice that the original function is even, you then know all behaviors are reflected across the $y$ axis. Little tricks like this can cut down on the amount of grunt work you have to do. :D