Answer:
[tex]\displaystyle (-\infty, -\frac{1}{6})[/tex]
Step-by-Step explanation:
We have the function:
[tex]\displaystyle y=\int_{1}^{x}\frac{1}{3+t+3t^2}\, dt[/tex]
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:
[tex]\displaystyle y^\prime=\frac{d}{dx}\Bigg[\int_{1}^{x}\frac{1}{3+t+3t^2}\, dt\Bigg][/tex]
By the Fundamental Theorem of Calculus:
[tex]\displaystyle y^\prime=\frac{1}{3+x+3x^2}[/tex]
So, we will take the derivative again. Hence:
[tex]\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][/tex]
We will use the chain rule. Let:
[tex]\displaystyle u=x^{-1}\text{ and } v=3+x+3x^2[/tex]
Differentiate:
[tex]\displaystyle y^\prime^\prime=-(3+x+3x^2)^{-2}(1+6x)[/tex]
Rewrite:
[tex]\displaystyle y^\prime^\prime=-\frac{6x+1}{(3+x+3x^2)^2}[/tex]
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:
[tex]\displaystyle 0=-\frac{6x+1}{(3+x+3x^2)^2}[/tex]
Multiplying both sides by the denominator gives:
[tex]0=-(6x+1)[/tex]
Then it follows that:
[tex]\displaystyle x=-\frac{1}{6}[/tex]
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:
[tex]\displaystyle y^\prime^\prime=-\frac{6(-1)+1}{3+(-1)+3(-1)^2}=1>0[/tex]
Since the result is positive, y is concave up for all values less than -1/6.
And testing for x=0, we see that:
[tex]\displaystyle y^\prime^\prime=-\frac{6(0)+1}{3+(0)+3(0)^2}=-\frac{1}{3}[/tex]
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:
[tex]\displaystyle (-\infty, -\frac{1}{6})[/tex]
Note that we use parentheses instead of brackets since at exactly x=-1/6, our graph is neither concave up nor concave down.
