Answer:
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 <em>x=-1/6. </em>
We will test for values less than and greater than this inflection point.
Testing for <em>x=-1</em>, we see that:
Since the result is positive, y is concave up for all values less than -1/6.
And testing for <em>x=0</em>, we see that:
Since the result is negative, y is concave down for all values greater than -1/6.
Therefore, the interval for which <em>y</em> is concave up is:
Note that we use parentheses instead of brackets since at exactly <em>x=-1/6</em>, our graph is neither concave up nor concave down.