Answer
(f + g)(t - 2) = $t^{2}$ - 3t + 3
Work Step by Step
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
