\( V \) \[ \begin{array}{l}\text { Suppose that the functions } r \text { and } s \text { are defined for all real numbers } x \text { as follows. } \\ r(x)=x+1 \\ s(x)=3 x^{2}\end{array} \] Whte the expressions for \( (r-s)(x) \) and \( (r+s)(x) \) and exaluate \( (r-s)(-1) \). \( \begin{array}{l}(r-s)(x)=\square \\ (r+s)(x)=\square \\ (r-s)(-1)=\square\end{array} \)

Let's determine the expressions for \((r - s)(x)\) and \((r + s)(x)\) given the functions \( r(x) = x + 1 \) and \( s(x) = 3x^2 \). Then, we'll evaluate \((r - s)(-1)\). ### 1. Finding \((r - s)(x)\) \[ (r - s)(x) = r(x) - s(x) = (x + 1) - 3x^2 = -3x^2 + x + 1 \] ### 2. Finding \((r + s)(x)\) \[ (r + s)(x) = r(x) + s(x) = (x + 1) + 3x^2 = 3x^2 + x + 1 \] ### 3. Evaluating \((r - s)(-1)\) Plug \( x = -1 \) into \((r - s)(x)\): \[ (r - s)(-1) = -3(-1)^2 + (-1) + 1 = -3(1) - 1 + 1 = -3 \] ### Summary \[ \begin{array}{l} (r - s)(x) = -3x^2 + x + 1 \\ (r + s)(x) = 3x^2 + x + 1 \\ (r - s)(-1) = -3 \end{array} \]