91Ó°ÊÓ

Polynomial functions are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. Generally, they are in the form of \[ P(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 \] For example, in the exercise: \[ f(x) = 3x^2 - x \] Here:

• 3 is a coefficient for \(x^2\)
• -1 is the coefficient for \(x\)
• The power of \(x\) in the second term is 1.
• there is no constant term or \(x^0\)

These functions can be graphically represented as curves, with the degree of the polynomial (highest power of \(x\)) affecting the shape of the curve. A degree 2 polynomial, like \(f(x)\) in this exercise, forms a parabola.

Function operations include various ways to combine or manipulate functions. Some common operations are:

• Addition: \((f+g)(x) = f(x) + g(x)\)
• Subtraction: \((f-g)(x) = f(x) - g(x)\)
• Multiplication: \((fg)(x) = f(x)g(x)\)
• Division: \((f/g)(x) = f(x)/g(x)\), as long as \ (g(x) eq 0)\

In this exercise, we are performing a subtraction of two functions: \[ f(x+1) - f(x) \] Which means we first calculate \(f(x+1)\), then subtract \( f(x) \) from it. This requires us to substitute \(x+1\) into \(f(x)\), which brings us to the next section.

Let’s understand substitution by focusing on the example from the exercise. To find \(f(x+1)\), we need to substitute \(x+1\) into the function \(f(x) = 3x^2 - x \). So,\[ f(x+1) = 3(x+1)^2 - (x+1) \] Let's break it down step by step. First we expand \((x+1)^2\): \[ (x+1)^2 = x^2 + 2x + 1 \] Substituting back in, we get: \[ f(x+1) = 3(x^2 + 2x + 1) - (x + 1) \] Simplifying further by distributing and combining like terms:\[ f(x+1) = 3x^2 + 6x + 3 - x - 1 \] \[ f(x+1) = 3x^2 + 5x + 2 \] Now let's find \( f(x+1) - f(x) \): \[ f(x+1) - f(x) = (3x^2 + 5x + 2) - (3x^2 - x) \] Notice that in subtraction many terms will cancel out: \[ f(x+1) - f(x) = 3x^2 + 5x + 2 - 3x^2 + x \] \[ f(x+1) - f(x) = 6x + 2 \] This shows the result of our function operation and helps in understanding how substitution and simplification are essential to solving such problems.