Need help and explain please!!

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.

not enough information

step-by-step explanation:

we cannot know this, because the other students could have handed out 10 flyers, or they could have handed out 100. one way, the mean would go up, and another way, the mean would go down.

87%

step-by-step explanation:

apex answer