Problem 2 - Applied Calculus

Question

If \(f(x)=x^{2}+1\), find and simplify: (a) \(f(t+1)\) (b) \(f\left(t^{2}+1\right)\) (c) \(f(2)\) (d) \(2 f(t)\) (e) \([f(t)]^{2}+1\)

Step-by-Step Solution

Answer

(a) \(t^2 + 2t + 2\), (b) \(t^4 + 2t^2 + 2\), (c) 5, (d) \(2t^2 + 2\), (e) \(t^4 + 2t^2 + 2\).

Step 1: Understanding the Function

We start with the function given as \( f(x) = x^2 + 1 \). This is the base function that we will use in all parts of the question.

Step 2: Substitute for f(t+1)

To find \( f(t+1) \), substitute \( t+1 \) into the function: \( f(t+1) = (t+1)^2 + 1 \). Expand the square: \( (t+1)^2 = t^2 + 2t + 1 \). Hence, \( f(t+1) = t^2 + 2t + 1 + 1 = t^2 + 2t + 2 \).

Step 3: Substitute for f(t^2+1)

To find \( f(t^2+1) \), substitute \( t^2+1 \) into the function: \( f(t^2+1) = (t^2+1)^2 + 1 \). Expand this as \( (t^2+1)^2 = t^4 + 2t^2 + 1 \). Hence, \( f(t^2+1) = t^4 + 2t^2 + 1 + 1 = t^4 + 2t^2 + 2 \).

Step 4: Evaluate f(2)

To find \( f(2) \), substitute \( 2 \) into the function: \( f(2) = 2^2 + 1 = 4 + 1 = 5 \).

Step 5: Calculate 2*f(t)

For \( 2f(t) \), first find \( f(t) = t^2 + 1 \). Multiply by 2: \( 2f(t) = 2(t^2 + 1) = 2t^2 + 2 \).

Step 6: Find [f(t)]^2 + 1

Calculate \([f(t)]^2 + 1\) by first finding \( f(t) = t^2 + 1 \) such that \([f(t)]^2 = (t^2 + 1)^2 = t^4 + 2t^2 + 1 \). Thus, \([f(t)]^2 + 1 = t^4 + 2t^2 + 1 + 1 = t^4 + 2t^2 + 2 \).

Key Concepts

Polynomial Functions

Polynomial functions are expressions involving variables with non-negative integer exponents, typically written in the form of a sum of terms. In the exercise, the function is given as \( f(x) = x^2 + 1 \). This is a simple polynomial function composed of two terms:

• The \( x^2 \) term which is quadratic.
• The constant term, \( +1 \).

Polynomial functions, like other mathematical functions, allow you to replace the variable \( x \) with different numbers, which fundamentally alters the output. The given function is quadratic since the highest power of \( x \) is 2. The graph of a quadratic polynomial looks like a U-shaped curve, which is symmetric about the y-axis if it's in the form \( x^2 + c \). In general, polynomial functions can have multiple terms and varying coefficients, which affect the shape and direction of their graph.

Function Evaluation

Function evaluation is the process of substituting specific values for the variable in a given function. In our exercise, we are asked to perform evaluation for several substitutions:

• For \( f(t+1) \), substitute \( t+1 \) in place of \( x \) in \( f(x) \), resulting in \( (t+1)^2 + 1 \).
• To find \( f(t^2+1) \), substitute \( t^2+1 \) for \( x \), getting \( (t^2+1)^2 + 1 \).
• Evaluating \( f(2) \) is simpler where you substitute \( 2 \), leading to \( 2^2 + 1 \).

This substitution process allows us to explore how the function behaves with different inputs. Essentially, by plugging values or expressions into the function, you see how the dependent variable changes, denoted by the result of each evaluation.

Algebraic Manipulation

Algebraic manipulation involves using algebraic rules and operations to rearrange or simplify mathematical expressions. In this exercise, manipulation helps simplify results obtained from function evaluation. Let's consider:

• For \( f(t+1) = (t+1)^2 + 1 \), expand \( (t+1)^2 \) to \( t^2 + 2t + 1 \) and simplify the expression to \( t^2 + 2t + 2 \).
• When calculating \( f(t^2+1) = (t^2+1)^2 + 1 \), first expand \( (t^2+1)^2 = t^4 + 2t^2 + 1 \) then add \( 1 \) to get \( t^4 + 2t^2 + 2 \).
• For \( 2f(t) \), multiply \( f(t) = t^2 + 1 \) by 2, resulting in \( 2t^2 + 2 \).

These steps involve understanding the distributive property and the method of combining like terms. By manipulating algebraically, you simplify expressions, making them easier to interpret and solve, an essential skill in algebra.