Chapter 1: Problem 2
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) \([f(t)]^{2}+1\)
Short Answer
Expert verified
(a) \(t^2 + 2t + 2\), (b) \(t^4 + 2t^2 + 2\), (c) 5, (d) \(2t^2 + 2\), (e) \(t^4 + 2t^2 + 2\).
Step by step solution
01
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.
02
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, \( f(t+1) = t^2 + 2t + 1 + 1 = t^2 + 2t + 2 \).
03
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, \( f(t^2+1) = t^4 + 2t^2 + 1 + 1 = t^4 + 2t^2 + 2 \).
04
Evaluate f(2)
To find \( f(2) \), substitute \( 2 \) into the function: \( f(2) = 2^2 + 1 = 4 + 1 = 5 \).
05
Calculate 2*f(t)
For \( 2f(t) \), first find \( f(t) = t^2 + 1 \). Multiply by 2: \( 2f(t) = 2(t^2 + 1) = 2t^2 + 2 \).
06
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, \([f(t)]^2 + 1 = t^4 + 2t^2 + 1 + 1 = t^4 + 2t^2 + 2 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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, \( +1 \).
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 \( 2^2 + 1 \).
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 \( 2t^2 + 2 \).