Answer:
Step-by-step explanation:
We are given the second derivative:
And we want to find its inflection points.
To do so, we will first determine possible inflection points. These occur whenever g''(x) = 0 or is undefined.
Next, we will test values for the intervals. Inflection points occur if and only if the sign changes before and after the point.
So first, finding the zeros, we see that:
So, we can draw the following number-line:
<----(-4)--------------(3)----(6)---->
Now, we will test values for the intervals x < -4, -4 < x < 3, 3 < x < 6, and x > 6.
Testing for x < -4, we can use -5. So:
Since we acquired a positive result, g(x) is concave up for x < -4.
For -4 < x < 3, we can use 0. So:
Since we acquired a negative result, g(x) is concave down for -4 < x < 3.
And since the sign changed before and after the possible inflection point at x = -4, x = -4 is indeed an inflection point.
For 3 < x < 6, we can use 4. So:
Since we acquired a negative result, g(x) is concave down for 3 < x < 6.
Since the sign didn't change before and after the possible inflection point at x = 3 (it stayed negative both times), x = -3 is not a inflection point.
And finally, for x > 6, we can use 7. So:
So, g(x) is concave up for x > 6.
And since we changed signs before and after the inflection point at x = 6, x = 6 is indeed an inflection point.
