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In mathematics, a function defines a relationship between a set of inputs and a set of permissible outputs, with each input being related to exactly one output. The basic idea is that if you have an input value (often called 'x'), the function will transform it into an output value through a specific rule. For example, in the function defined by the equation \(h(x) = \frac{x^2}{1-2x}\), for every input value of \(x\), there is a unique output value.Functions are an essential part of calculus and many other parts of mathematics because they provide a clear rule for taking an input and calculating an output. This makes them versatile tools in mathematical modeling. They are often expressed using a formula that describes precisely what happens to an input when it is used in the function.

A rational function is a type of function that can be expressed as the ratio of two polynomials. The general form of a rational function is \(f(x) = \frac{P(x)}{Q(x)}\) where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x) eq 0\). Rational functions are interesting because they can have vertical asymptotes, horizontal asymptotes, and holes, making their graphs more complicated than polynomial functions.For the function \(h(x) = \frac{x^2}{1-2x}\), it is clear that it is a rational function. Here, \(x^2\) is the numerator and \(1-2x\) is the denominator. One important aspect of rational functions is that they are not defined when the denominator is zero, as division by zero is undefined in mathematics. In the case of \(h(x)\), the function is undefined for \(x = \frac{1}{2}\), as it would make the denominator zero.

Function evaluation is the process of substituting a specific value into a function and simplifying to find the result. When evaluating functions, you essentially follow the instructions given by the function’s formula with the chosen input value.For instance, to evaluate \(h(0)\) from the function \(h(x) = \frac{x^2}{1-2x}\), you substitute \(x = 0\) into the formula. This gives \(h(0) = \frac{0^2}{1-2 \times 0} = \frac{0}{1} = 0\). You simply execute the arithmetic to find the output. Function evaluation is a fundamental skill in calculus, as it involves applying specific inputs to explore and understand the behavior and outputs of functions.

Substitution in functions involves replacing the variable in a function with a specific value or another expression. This method is vital for determining how a function responds to different inputs and is part of evaluating the function.In the given exercise, various substitutions are made:

• For \(h(3)\), replace \(x\) with 3 to get \(h(3) = \frac{3^2}{1 - 2 \times 3} = \frac{9}{-5} = -\frac{9}{5}\).
• For \(h(p+1)\), replace \(x\) with \(p+1\), which involves more algebra: \(h(p+1) = \frac{(p+1)^2}{1 - 2(p+1)} = \frac{p^2 + 2p + 1}{-2p - 1}\).

Substitution allows us to go beyond numerical inputs, letting us see the functional behavior with variable inputs or parameterized expressions. It's a key part of exploring functions analytically.