13. suppose ( g(x)=2 - x ).\na. on what intervals is ( g ) concave up and on what intervals is ( g ) concave…

Explanation:

Step1: Find critical points

Set (g''(x)=0), so (2 - x = 0), which gives (x = 2).

Step2: Test intervals

• For (x<2), let (x = 0). Then (g''(0)=2-0 = 2>0).
• For (x>2), let (x = 3). Then (g''(3)=2 - 3=-1<0).

Since (g''(x)>0) when (x<2), (g(x)) is concave up on the interval ((-\infty,2)). Since (g''(x)<0) when (x>2), (g(x)) is concave down on the interval ((2,\infty)).

Step3: Find inflection point

The inflection point occurs where (g''(x)) changes sign. We know (x = 2). To find the (y) - coordinate of the inflection point, we need to integrate (g''(x)) twice (but since we are only asked for the (x) - value of the inflection point based on the second - derivative test, and the second - derivative changes sign at (x = 2)). The inflection point is at (x = 2).

Answer:

a. (g(x)) is concave up on ((-\infty,2)) and concave down on ((2,\infty)). b. The inflection point of (g(x)) is at (x = 2).