91Ó°ÊÓ

Understanding the symmetry of functions is essential in determining whether a function is even or odd. A function is symmetric with respect to the y-axis if, for every point \( (x, y) \) on the function, there is a corresponding point \( (-x, y) \). If this is true for all points on the function, then the function is considered even. On the other hand, a function exhibits symmetry with respect to the origin if, for each point \( (x, y) \) on the function, there is a corresponding point \( (-x, -y) \). If a function meets this criterion for all points, it is described as odd.

To check if a function is even, we evaluate \( f(x) \) and \( f(-x) \). If \( f(x) = f(-x) \), the function is even. For odd functions, we look for the condition \( f(-x) = -f(x) \). In our original exercise, we found that \( g(s) = 4 s^{2 / 3} \) yielded the same result when \( s \) was replaced by \( -s \) which implies that \( g(s) \) is symmetric with respect to the y-axis and therefore even.

Power functions have the form \( f(x) = ax^n \) where \( a \) and \( n \) are real numbers and \( n \) is typically termed the power or exponent. The value of \( n \) plays a crucial role in determining the symmetry properties of the function. For instance, if \( n \) is an even integer, the power function is even, and if \( n \) is an odd integer, the power function is odd.

In our example, the function \( g(s) = 4s^{2 / 3} \) is a power function with \( n \) as a fraction. When dealing with fractional powers, the even or odd nature can be deduced by examining the numerator of the exponent. Since \( 2 \) (the numerator) is even, the function behaves similarly to an even power function over the domain where it is defined. \( g(s) \) retains its value when \( s \) is replaced with \( -s \) indicating it is an even function.

Evaluating a function is a fundamental skill in mathematics where we find the output of a function given an input. When we assess whether a function is even or odd, we use function evaluation at \( x \) and \( -x \) to compare results. This assessment was key in the exercise where we evaluated \( g(s) \) and \( g(-s) \) to determine the symmetry of the function.

In our step-by-step solution, evaluating at \( s \) and \( -s \) ensured that we compared the corresponding outputs, \( g_1 \) and \( g_2 \). By determining that \( g_1 = g(s) = 4 s^{2 / 3} \) and \( g_2 = g(-s) = 4(-s)^{2 / 3} = 4s^{2 / 3} \) are equivalent, we established that \( g(s) \) is indeed an even function. It’s crucial for students to practice this technique, as function evaluation lays the groundwork for understanding more complex concepts in algebra and calculus.