91Ó°ÊÓ

For condition a, the function satisfies that at \(x = 2\), \(g(2) = 2\), meaning the graph passes through the point \((2, 2)\). It also describes how the derivatives behave around \(x = 2\). - For \(x < 2\): The derivative \(g'(x)\) is between 0 and 1, meaning the function is increasing moderately. - As \(x\) approaches 2 from the left \((x \rightarrow 2^{-})\), the derivative \(g'(x)\) approaches 1 from the left, indicating the slope of the tangent becomes more positive and approaches exactly 1 as it reaches 2.- For \(x > 2\): The derivative becomes negative and falls within the range \(-1 < g'(x) < 0\), which means the function is decreasing for \(x > 2\). - As \(x \rightarrow 2^{+}\), \(g'(x)\) approaches -1 from the positive direction, indicating a decreasing slope that becomes -1 exactly at \(x = 2\).

Based on the analysis above, sketch a graph that increases with a slope nearing 1 as it approaches \(x = 2\) from the left and decreases with a slope approaching -1 as it moves to the right of \(x = 2\). At \(x = 2\), the point \((2, 2)\) acts as a critical point where the nature of the slope changes from increasing to decreasing.

For condition b, also at \(x = 2\), \(g(2) = 2\). The behavior of derivatives around 2 is different:- For \(x < 2\): The derivative is negative \(g'(x) < 0\), meaning the function is decreasing.- As \(x\) approaches 2 from the left, \(g'(x)\) approaches \(-\infty\), indicating that the slope becomes steeply negative immediately before \(x = 2\).- For \(x > 2\): The derivative \(g'(x) > 0\), meaning the function increases for \(x > 2\).- As \(x\) approaches 2 from the right, \(g'(x)\) approaches \(\infty\), indicating the slope becomes steeply positive immediately after \(x = 2\).

Sketch a graph that decreases with a slope becoming increasingly steeply negative as \(x\) approaches 2 from the left, and increases with a steeply positive slope as \(x\) moves away from 2 on the right. This results in a sharp 'V' shape at the point \((2,2)\).