f(t+1) = 3(t^2 + 2t + 1) + 5(t+1) + 2 - Dyverse
Understanding and Simplifying the Function: f(t+1) = 3(t² + 2t + 1) + 5(t+1) + 2
Understanding and Simplifying the Function: f(t+1) = 3(t² + 2t + 1) + 5(t+1) + 2
Mathematics often presents functions in complex forms that may seem intimidating at first glance, but breaking them down clearly reveals their structure and utility. In this article, we’ll explore the function equation:
f(t+1) = 3(t² + 2t + 1) + 5(t+1) + 2
Understanding the Context
We’ll simplify it step-by-step, interpret its components, and explain how this function behaves—and why simplifying helps in solving equations, graphing, or applying it in real-world contexts.
Step 1: Simplify the Right-Hand Side
We begin by expanding and combining like terms on the right-hand side of the equation:
Key Insights
f(t+1) = 3(t² + 2t + 1) + 5(t + 1) + 2
Step 1.1 Expand each term:
- 3(t² + 2t + 1) = 3t² + 6t + 3
- 5(t + 1) = 5t + 5
Step 1.2 Add all expanded expressions:
f(t+1) = (3t² + 6t + 3) + (5t + 5) + 2
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Step 1.3 Combine like terms:
- Quadratic: 3t²
- Linear: 6t + 5t = 11t
- Constants: 3 + 5 + 2 = 10
So,
f(t+1) = 3t² + 11t + 10
Step 2: Understand What f(t+1) Means
We now have:
f(t+1) = 3t² + 11t + 10
This form shows f in terms of (t + 1). To find f(u), substitute u = t + 1, which implies t = u − 1.
