Final answer:
To find the global maximum and minimum of the function f(t) = 3t/(1+t²), the derivative f'(t) was found, set to zero to locate critical points, which are at t = ±1. After plugging these points back into f(t), the global maximum was determined to be 1.5, and the global minimum -1.5.
Step-by-step explanation:
To find the global maximum and minimum values of the function f(t) = 3t/(1+t²), we must consider both the critical points and the behavior at infinity.
Step 1: Find the derivative
First, find the derivative f'(t) to locate the critical points. Using the quotient rule:
f'(t) = (3(1+t²) - 3t(2t))/ (1+t²)^2 = (3 - 3t²) / (1+t²)^2.
Step 2: Find the critical points
Then, set this derivative equal to zero to find critical points.
0 = (3 - 3t²) / (1+t²)^2 implies that t² = 1, so t = ±1.
Step 3: Behavior at infinity
Since the degree of the denominator is greater than the numerator, as t approaches ± infinity, f(t) approaches zero.
Step 4: Determine maximum and minimum values
Plugging the critical points into f(t), we get: f(1) = 1.5 and f(-1) = -1.5. Checking our end behavior analysis, we see there are no larger or smaller values as t approaches infinity, so our global maximum is 1.5 and our global minimum is -1.5.