I need help. Is it possible to do this without graphing?

Step-by-step explanation:

We are given the second derivative:

And we want to find the values for -3 < x < 4 in which the graph of g has a point of inflection where the concavity changes from concave up to down.

First, recall that inflection points occur whenever g''(x) = 0 or is undefined.

And that:

g is concave up for an interval if g''(x) > 0 for that interval. g is concave down for an interval if g''(x) < 0 for that interval.

Graphing the function (please refer below for the graph), we can see that g''(x) has an inflection point at x = -2.6, 0, 2.3, and 3.2.

If g''(x) changes from concave up to concave down, the values before and after that inflection point must be positive and then negative.

We can see that g''(x) is only positive for the values before the inflection points at x = 0 and x = 2.3.

However, for values after the inflection points, only x = 2.3 becomes negative.

Therefore, since at x = 2.3, the graph changes from positive to negative, this is where the graph of g changes from concave up to concave down.

Notes:

For the point x = 0, the graph stayed positive both before and after x = 0.

Therefore, g does not have an inflection point at x = 0.

Ido not know i use to

Okay so for every triangle has three angles that always add up to 180 degrees. and to tell the difference between whether they are secure, right or obtuse is that acute: has one angle that is less than 90 degrees. right triangle means there is at least one angle that is exactly 90 degrees. and an obtuse angle has an angle that is bigger than 180 degrees. i’ll do #13 as an example: 13. 40 and 50 so if every triangle has 180 degrees 40+50= 90 degrees then180-90= 90 degrees so the third angle is 90 degrees and because this triangle has a 90 degree angle it means it is a right triangle