Question

Using words and figures, explain why the range of \(f(x)=x^{n},\) where \(n\) is a positive odd integer, is all real numbers. Explain why the range of \(g(x)=x^{n},\) where \(n\) is a positive even integer, is all nonnegative real numbers.

Short answer

Question: Explain why the range of \(f(x) = x^n\), where \(n\) is a positive odd integer, is all real numbers and why the range of \(g(x) = x^n\), where \(n\) is a positive even integer, is all nonnegative real numbers. Answer: The range of \(f(x) = x^n\) is all real numbers for positive odd \(n\) because it is an odd function, meaning it has rotational symmetry with respect to the origin; hence, if \(x^n\) has certain positive values, it also has corresponding negative values. The range of \(g(x) = x^n\) is all nonnegative real numbers for positive even \(n\) because it is an even function, meaning it has reflectional symmetry about the y-axis; thus, if \(x^n\) has positive values, their corresponding negative values also have the same positive value.

Step-by-step solution

Define odd and even functions

An odd function is a function that satisfies the property \(f(-x) = -f(x)\) for all x in the domain. An even function is a function that satisfies the property \(f(-x) = f(x)\) for all x in the domain.

Check the properties of \(f(x)=x^n\) with odd \(n\)

To check if the function \(f(x) = x^n\) is an odd function when \(n\) is a positive odd integer, we need to check whether \(f(-x) = -f(x)\) holds true. Let \(n = 2k+1\) where \(k\) is a non-negative integer (odd can be represented in this form). Then, \(f(-x) = (-x)^{2k+1} = -x^{2k+1} = -f(x)\). Since this equality holds true for all \(x\), we can conclude that \(f(x) = x^n\) is an odd function when \(n\) is a positive odd integer.

Determine the range of \(f(x)=x^n\) with odd \(n\)

Since \(f(x)\) is an odd function, it has rotational symmetry with respect to the origin. This means that for every positive \(y\) value, there will always be a corresponding negative \(y\) value and vice versa. In other words, if \(x^n\) has certain positive values, it will also have their corresponding negative values. Hence, the range of \(f(x) = x^n\) for positive odd \(n\) will be all real numbers.

Check the properties of \(g(x)=x^n\) with even \(n\)

To check if the function \(g(x) = x^n\) is an even function when \(n\) is a positive even integer, we need to check if \(g(-x) = g(x)\) holds true. Let \(n = 2k\) where \(k\) is a positive integer (even can be represented in this form). Then, \(g(-x) = (-x)^{2k} = x^{2k} = g(x)\). Since this equality holds true for all \(x\), we can conclude that \(g(x) = x^n\) is an even function when \(n\) is a positive even integer.

Determine the range of \(g(x)=x^n\) with even \(n\)

Since \(g(x)\) is an even function, it has reflectional symmetry about the y-axis. This means that for every positive \(x\) value, there will always be a corresponding negative \(x\) value with the same \(y\) value. In other words, if \(x^n\) has certain positive values, their corresponding negative values will also have the same positive value, as the function is even. Thus, the range of \(g(x) = x^n\) for positive even \(n\) will be all nonnegative real numbers. In conclusion: - The range of \(f(x) = x^n\), where \(n\) is a positive odd integer, is all real numbers. - The range of \(g(x) = x^n\), where \(n\) is a positive even integer, is all nonnegative real numbers.