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

`\displaystyle (-\infty, -(1)/(6))`

Explanation:

We have the function:

`\displaystyle y=\int_(1)^(x)(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=(d)/(dx)\Bigg[\int_(1)^(x)(1)/(3+t+3t^2)\, dt\Bigg]`

By the Fundamental Theorem of Calculus:

`\displaystyle y^\prime=(1)/(3+x+3x^2)`

So, we will take the derivative again. Hence:

`\displaystyle y^\prime^\prime=(d)/(dx)\Big[(1)/(3+x+3x^2)\Big]=(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=-(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=-(6x+1)/((3+x+3x^2)^2)`

Multiplying both sides by the denominator gives:

`0=-(6x+1)`

Then it follows that:

`\displaystyle x=-(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=-(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=-(6(0)+1)/(3+(0)+3(0)^2)=-(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, -(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.