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

Step-by-Step explanation:

We have the function:

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:

By the Fundamental Theorem of Calculus:

So, we will take the derivative again. Hence:

We will use the chain rule. Let:

Differentiate:

Rewrite:

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:

Multiplying both sides by the denominator gives:

Then it follows that:

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:

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

And testing for x=0, we see that:

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:

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

i think is 3 and 6. this took me 10 min to think.hope to and hope that my answer is brainliest.(follow me )

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step-by-step explanation: