Find two real numbers such that f does not change concavity

[agilder23]

Here's a problem. I know how to do one similar to this one, but it didn't involve concavity.

Let f(x)=(11x^2+1)/(7x^2+2x+7). Find two real numbers a and b such that f(x) does not change its concavity on (-infinity, a) and on (b, infinity).

Any suggestions as to where to begin?

 

[soroban]

Hello, agilder23!

I have a game plan . . . but I hope Someone Else will do the work.

\(\displaystyle \text{Let: }\:f(x)\:=\:\frac{11x^2+1}{7x^2+2x+7}\)

\(\displaystyle \text{Find two real numbers }a\text{ and }b\text{ such that:}\)

. . \(\displaystyle f(x)\text{ does not change its concavity on }(-\infty, a)\text{ and on }(b, \infty)\)

We have two tasks . . . each worthy of Hercules . . .

[1] Crank out its second derivative and equate to zero: .\(\displaystyle f''(x) \:=\:0\)

[2] Solve and determine its inflection points.


Here's my reasoning:

I believe we have a cubic to solve, so there are three inflection points:
. . \(\displaystyle x\:=\: p,q,r\;\;\text{ with }\ < q < r.\)

These are the values at which the concavity changes.

\(\displaystyle \text{Hence, the concavity does }not\text{ change on: }\-\infty,p)\:\text{ and }\r,\infty)\)

 

[BigGlenntheHeavy]

Whipping out my trusty TI-89, the two points of inflection are

(-.583246,.577247) and (.576582,.44435).

Hence f(x) is concave down on the interval (-?, -.583246]µ[.576582,?) and concave up on the interval [-.583246,.576582].

 

[BigGlenntheHeavy]

A revision on this problem.

f(x) = (11x²+1)/(7x²+2x+7), hence d²y/dx² = -4(77x³+735x²-21x-247)/(7x²+2x+7)³.

Therefore, setting the second derivative = to zero, we get three points of inflection, which are:

(-9.53879,1.6034), the one I missed), (-.58324,.57724), and (.576582,.44435). Soroban was right, my apologies.