Problem 6 - Applied Calculus

Question

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

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: Find \( f(t+1) \)

To find \( f(t+1) \), replace \( x \) with \( t+1 \) in the expression for \( f(x) \). So, start with \( f(x) = x^2 + 1 \).\[ f(t+1) = (t+1)^2 + 1 \].Next, expand \((t+1)^2\) to get \(t^2 + 2t + 1\).Thus, \[ f(t+1) = t^2 + 2t + 1 + 1 = t^2 + 2t + 2 \].

Step 2: Find \( f(t^2 + 1) \)

Replace \( x \) with \( t^2 + 1 \) in the expression for \( f(x) = x^2 + 1 \).\[ f(t^2 + 1) = (t^2 + 1)^2 + 1 \].Expand \((t^2 + 1)^2\):\[ (t^2 + 1)^2 = t^4 + 2t^2 + 1 \].So,\[ f(t^2 + 1) = t^4 + 2t^2 + 1 + 1 = t^4 + 2t^2 + 2 \].

Step 3: Find \( f(2) \)

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

Step 4: Find \( 2f(t) \)

First, find \( f(t) \) which is straight from the function, \( f(t) = t^2 + 1 \).Now, multiply by 2:\[ 2f(t) = 2(t^2 + 1) = 2t^2 + 2 \].

Step 5: Find \( (f(t))^2 + 1 \)

Start with \( f(t) = t^2 + 1 \) and square it:\[ (f(t))^2 = (t^2 + 1)^2 \], which expands to:\[ t^4 + 2t^2 + 1 \].Now add 1:\[ (f(t))^2 + 1 = t^4 + 2t^2 + 1 + 1 = t^4 + 2t^2 + 2 \].

Key Concepts

Quadratic Functions

Quadratic functions are a category of polynomial functions where the highest power of the variable is two. Written in the general form as \( f(x) = ax^2 + bx + c \), these functions describe parabolas when graphed on the Cartesian plane.
For the function \( f(x) = x^2 + 1 \), the coefficients are \( a = 1 \), \( b = 0 \), and \( c = 1 \).
- The graph of this function is a parabola opening upwards because the coefficient of the \( x^2 \) term is positive.- It is a simple quadratic without the linear \( x \) term, making it symmetric with respect to the y-axis.
Evaluating a quadratic function, such as \( f(t+1) \), involves substituting a new expression in place of \( x \) and simplifying it. These evaluations often require expanding squared binomials, as seen when expanding \((t+1)^2\). This consists of applying the algebraic identity \((a + b)^2 = a^2 + 2ab + b^2\).
By understanding how to manipulate and substitute in quadratic expressions, students can solve a variety of problems involving these types of functions.

Polynomial Operations

Polynomial operations involve adding, subtracting, multiplying, and sometimes dividing polynomial expressions. A polynomial is defined by terms involving a variable raised to whole number powers where each term has a coefficient.
- When performing polynomial operations, it's important to follow the algebraic rules for simplification.- The terms need to be collected like \( (x^2 + 2x + 1 ) \) as seen while expanding \( (t + 1)^2 \).
In the original exercise, the multiplication operation is shown, for example as seen in \( 2f(t) \), which involves distributing the number 2 across the polynomial \( t^2 + 1 \) to yield \( 2t^2 + 2 \). A key step is the expansion of binomials using distributes and combines terms to simplify them. When extending this to more complex expressions, keep track of each step to avoid errors and ensure all like terms are combined perfectly.
Mastering operations on polynomials is foundational as it prepares students to tackle more complex algebraic expressions and functions.

Composition of Functions

The composition of functions refers to applying one function to the results of another, essentially feeding the output of one function as the input into another.
It is denoted as \( (f \circ g)(x) = f(g(x)) \). To carry out this process, understand which function is applied first and which is applied second.
In the given exercise, evaluating \( f(t^2 + 1) \) demonstrates this concept. Here, first, the function outputs of \( t^2 + 1 \) are treated as inputs into the function \( f \).
To solve, substitute \( t^2 + 1 \) into \( f(x) = x^2 + 1 \), i.e., find \( (t^2 + 1)^2 + 1 \)'. This involves both expansion and combining like terms, showcasing the nature of composition.

• First, tackle the inner function \( t^2 + 1 \).
• Then, use the result as the argument of the outer function \( f \).

Grasping function composition is a powerful tool that extends into calculus and various applications, making it a fundamental topic in understanding more advanced mathematics.