Question

I can't seem to figure this one out. Could someone show how to do it and the answer? Thanks in advance!

Answer 1

Answer:

$\displaystyle (-\infty, -\frac{1}{6})$

Step-by-Step explanation:

We have the function:

$\displaystyle y=\int_{1}^{x}\frac{1}{3+t+3t^2}\, dt$

And we want to find the interval for which y is concave upwards.

Therefore, we will need to find the second derivative of y, find its inflection points (where y''=0), and test for values.

So, let's take the derivative of both sides with respect to x. So:

$\displaystyle y^\prime=\frac{d}{dx}\Bigg[\int_{1}^{x}\frac{1}{3+t+3t^2}\, dt\Bigg]$

By the Fundamental Theorem of Calculus:

$\displaystyle y^\prime=\frac{1}{3+x+3x^2}$

So, we will take the derivative again. Hence:

$\displaystyle y^\prime^\prime=\frac{d}{dx}\Big[\frac{1}{3+x+3x^2}\Big]=\frac{d}{dx}\Big[(3+x+3x^2)^{-1}\Big]$

We will use the chain rule. Let:

$\displaystyle u=x^{-1}\text{ and } v=3+x+3x^2$

Differentiate:

$\displaystyle y^\prime^\prime=-(3+x+3x^2)^{-2}(1+6x)$

Rewrite:

$\displaystyle y^\prime^\prime=-\frac{6x+1}{(3+x+3x^2)^2}$

So, points of inflection, where the concavity changes, is whenever the second derivative is 0 or undefined.

We can see that the second derivative will never be undefined since the denominator can never equal 0.

So, our only possible inflection points are when it's equal to 0. Hence:

$\displaystyle 0=-\frac{6x+1}{(3+x+3x^2)^2}$

Multiplying both sides by the denominator gives:

$0=-(6x+1)$

Then it follows that:

$\displaystyle x=-\frac{1}{6}$

So, our only possible point of inflection is at x=-1/6.

We will test for values less than and greater than this inflection point.

Testing for x=-1, we see that:

$\displaystyle y^\prime^\prime=-\frac{6(-1)+1}{3+(-1)+3(-1)^2}=1>0$

Since the result is positive, y is concave up for all values less than -1/6.

And testing for x=0, we see that:

$\displaystyle y^\prime^\prime=-\frac{6(0)+1}{3+(0)+3(0)^2}=-\frac{1}{3}$

Since the result is negative, y is concave down for all values greater than -1/6.

Therefore, the interval for which y is concave up is:

$\displaystyle (-\infty, -\frac{1}{6})$

Note that we use parentheses instead of brackets since at exactly x=-1/6, our graph is neither concave up nor concave down.