ismahan's question at Yahoo! Answers regarding concavity

[MarkFL]

Here is the question:

Calculus Homework Please Help!!?

Let f(x)= 1/(7x^2+8). Find the open intervals on which f is concave up . Then determine the x-coordinates of all inflection points of f.

1. F is concave up on the intervals ______________
2. F is concave down on the intervals ________________
3. The inflection points occur at x= ____________

Notes: In the first two, your answer should either be a single interval, such as (0,1), a comma separated list of intervals, such as (-inf, 2), (3,4), or the word "none".
In the last one, your answer should be a comma separated list of x values or the word "none".

Here is a link to the question:

Calculus Homework Please Help!!? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello ismahan,

We are given the function:

$\displaystyle f(x)=\frac{1}{7x^2+8}=(7x^2+8)^{-1}$

We need to find the second derivative to answer questions concerning concavity. Using the power and chain rules, we find:

$\displaystyle f'(x)=(-1)(7x^2+8)^{-2}(14x)=-14x(7x^2+8)^{-2}$

Differentiate again, using the product, power and chain rules:

$\displaystyle f''(x)=(-14x)(-2(7x^2+8)^{-3}(14x))+(-14)(7x^2+8)^{-2}=\frac{14(21x^2-8)}{(7x^2+8)^3}$

Now, we observe that the denominator has no real roots, so our only critical values come from the numerator. Equating the factor containing $x$ to zero will find for us the values of $x$ at which the sign of the second derivative may change:

$\displaystyle 21x^2-8=0$

$\displaystyle x^2=\frac{8}{21}$

$\displaystyle x=\pm\frac{2\sqrt{2}}{\sqrt{21}}$

We have two critical values, so we will have 3 intervals that result when dividing the $x$-axis at these two values. Since the roots are of odd multiplcity, we know the sign of the second derivative will alternate across these intervals, so we need only check one interval, and for simplicity, I suggest checking the middle interval with the test value $x=0$. In doing so, we can easily see the sign of the second derivative is negative in this middle interval.

1.) Hence, the given function is concave up on the intervals:

$\displaystyle \left(-\infty,-\frac{2\sqrt{2}}{\sqrt{21}} \right),\,\left(\frac{2\sqrt{2}}{\sqrt{21}},\infty \right)$

2.) The function is concave down on the interval:

$\displaystyle \left(-\frac{2\sqrt{2}}{\sqrt{21}},\frac{2\sqrt{2}}{\sqrt{21}} \right)$

3.) The function has inflection points for:

$\displaystyle x=-\frac{2\sqrt{2}}{\sqrt{21}},\,\frac{2\sqrt{2}}{ \sqrt{21}}$