\(\mathrm{f}(\mathrm{x})=|\mathrm{x}+2|+|\mathrm{x}-1|\) is (a) increasing in \((-1, \infty)\) (b) increasing in \([-1, \infty]\) (c) decreasing in \((-\infty,-2)\) (d) decreasing in \([-\infty,-2]\)

Given that \(f(x) = |x+2| + |x-1|\), let's write the function as a piecewise function considering the expressions inside the absolute value. Let's define: \(g(x) = x + 2\) and \(h(x) = x - 1\) Now, we can write the piecewise function: \[ f(x) = \begin{cases} -g(x)-h(x) & x<-2 \\ -g(x)+h(x) & -2 \le x < 1 \\ g(x)+h(x) & x \ge 1 \end{cases} \]

Now, we will find the derivatives for each case. 1. For the first case, where \(x < -2\), the derivative is: \(f'(x) = -g'(x) - h'(x)\) 2. For the second case, where \(-2 \le x < 1\), the derivative is: \(f'(x) = -g'(x) + h'(x)\) 3. For the third case, where \(x \ge 1\), the derivative is: \(f'(x) = g'(x) + h'(x)\) Now, we can find the derivatives for \(g'(x)\) and \(h'(x)\): \(g'(x) = 1\) and \(h'(x) = 1\)

Now, let's analyze the signs of the derivatives for each case, which will let us know if the function is increasing or decreasing. 1. For the first case, where \(x < -2\), \(f'(x) = -1 - 1 = -2\). Since \(f'(x) < 0\), the function is decreasing. 2. For the second case, where \(-2 \le x < 1\), \(f'(x) = -1 + 1 = 0\). Since \(f'(x) = 0\), the function is neither increasing nor decreasing. 3. For the third case, where \(x \ge 1\), \(f'(x) = 1 + 1 = 2\). Since \(f'(x) > 0\), the function is increasing. Based on this analysis, we can conclude the correct answer from the given options. a) increasing in \((-1, \infty)\): This is not correct, because the function is only increasing for \(x \ge 1\). b) increasing in \([-1, \infty]\): This is not correct, because the function is only increasing for \(x \ge 1\). c) decreasing in \((-\infty, -2)\): This is the correct answer, as the function is decreasing for all values of \(x < -2\). d) decreasing in \([- \infty, -2]\): This is not correct, because the function is only decreasing for \(x < -2\).