91Ó°ÊÓ

Function evaluation refers to the process of substituting a specific value of the independent variable (usually represented as \(x\)) into a function to find the corresponding output. This is a fundamental skill in precalculus and calculus. Evaluating a function essentially tells us what the output of the function is when the input is known.

For example, if we have the function \(f(x)=\frac{10}{1+2 e^{-0.3 x}}\), and we want to determine the value when \(x = 0\), we substitute \(0\) in place of \(x\). This substitution gives us a new expression: \(f(0)=\frac{10}{1+2 e^{-0.3(0)}}\).

It’s like plugging in numbers into a formula to see what comes out on the other side. This involves ensuring all operations within the function are correctly applied to the substituted value.

Exponential functions are a crucial component of mathematics, often written in the form \(a \, e^{kx}\), where \(e\) is the mathematical constant approximately equal to 2.718, and \(a\) and \(k\) are constants. These functions model growth or decay processes and are notable for their rapid increase or decrease.

In the exercise, \(e^{-0.3x}\) is part of the denominator in the function. Substituting \(x = 0\), we focus on simplifying \(e^{-0.3(0)}\). Utilizing the important exponential rule that \(e^0 = 1\) (since any non-zero number raised to the power of zero equals one), our expression simplifies to \(1\).

Therefore, when calculating \(f(0)\), the expression simplifies to \(f(0) = \frac{10}{1+2(1)}\). Exponential functions often seem challenging, but they follow straightforward rules, especially when \(x\) equals zero or one.

Simplifying expressions involves reducing a mathematical expression into its simplest form. This makes the expression easier to work with, and it often requires a combination of arithmetic and algebraic methods.

After substituting the value for \(x\) in function evaluation and simplifying the exponential part, we are often left with a simpler arithmetic problem. For example, in \(\frac{10}{1+2 * 1}\), the exponential simplification of \(e^0\) (which is \(1\)) means our function evaluation of \(f(0)\) becomes \(\frac{10}{3}\).

Here, simplifying means adding and multiplying the numbers in the denominator first: \(1 + 2 * 1 = 3\). It's always crucial to perform operations in the correct sequence: simplify terms inside brackets, resolve exponents, perform multiplication or division, and finally, addition or subtraction. This methodical approach helps ensure accuracy and clarity in mathematics.