Question

1. suppose ( g(x)=2 - x ).

a. on what intervals is ( g ) concave up and on what intervals is ( g ) concave down?
b. state the inflection points of ( g ).

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 \(x = 2\).