Algebra and Trigonometry 10th Edition

(f + g)(t - 2) = $t^{2}$ - 3t + 3
The notation (f + g)(t - 2) can be rewritten as f(t - 2) + g(t - 2). Solving by adding the functions then evaluating at t - 2 will yield the same result as evaluating each function and then adding together. In this question: f(x) = x + 3 g(x) = $x^{2}$ - 2 Method 1: Add the functions then evaluate First we want to add the two functions. The new function will be called h(x). h(x) = f(x) + g(x) h(x) = (x + 3) + ($x^{2}$ - 2) h(x) = x + 3 + $x^{2}$ - 2 Combine like variables to get: h(x) = $x^{2}$ + x + 1 Evaluate at x = t - 2: h(t - 2) = $(t - 2)^{2}$ + (t - 2) + 1 = $t^{2}$ - 4t + 4 + t - 2 + 1 = $t^{2}$ - 3t + 3 Method 2: Evaluate the functions and then add First we want to evaluate f(x) at x = t - 2: f(t - 2) = t - 2 + 3 = t + 1 Then we want to evaluate g(x) at x = t - 2: g(t - 2) = $( t - 2)^{2}$ - 2 = $t^{2}$ - 4t + 4 - 2 = $t^{2}$ - 4t + 2 Add the numbers together: (f + g)(t - 2) = t + 1 + $t^{2}$ - 4t + 2 = $t^{2}$ - 3t + 3 Using both methods (f + g)(t - 2) = $t^{2}$ - 3t + 3