Question

what are the domain and range of the function?
$f(x) = \frac{3}{5x^5}$
domain: $(-\infty, \infty)$
\\(\circ\\) range: $(-\infty, 0)$

domain: $(-\infty, 0) \cup (0, \infty)$
\\(\circ\\) range: $(-\infty, 0) \cup (0, \infty)$

domain: $(-\infty, 0) \cup (0, \infty)$
\\(\circ\\) range: $(-\infty, 0)$

domain: $(-\infty, 0) \cup (0, \infty)$
\\(\circ\\) range: $(0, \infty)$

Explanation:

Step1: Find the domain

The function is \( f(x)=\frac{3}{5x^{5}} \). A fraction is undefined when the denominator is zero. So, we set the denominator equal to zero and solve for \( x \):
\( 5x^{5}=0 \)
Divide both sides by 5: \( x^{5}=0 \)
Take the fifth root of both sides: \( x = 0 \)
So, the function is undefined at \( x = 0 \). Therefore, the domain is all real numbers except \( x = 0 \), which is \( (-\infty, 0)\cup(0, \infty) \).

Step2: Find the range

Let \( y=\frac{3}{5x^{5}} \). We can solve for \( x \) in terms of \( y \) to find the range.
First, multiply both sides by \( 5x^{5} \): \( y\cdot5x^{5}=3 \)
Then, divide both sides by \( 5y \) (assuming \( y
eq0 \)): \( x^{5}=\frac{3}{5y} \)
Take the fifth root of both sides: \( x=\sqrt[5]{\frac{3}{5y}} \)
For \( x \) to be a real number, the expression inside the fifth root must be a real number. Since the fifth root is defined for all real numbers, the only restriction is \( y
eq0 \) (because if \( y = 0 \), we would have division by zero in the step \( x^{5}=\frac{3}{5y} \)). Also, we can analyze the sign:

• If \( x>0 \), then \( x^{5}>0 \), so \( \frac{3}{5x^{5}}>0 \) (since 3 and \( 5x^{5} \) are positive).
• If \( x<0 \), then \( x^{5}<0 \), so \( \frac{3}{5x^{5}}<0 \) (since 3 is positive and \( 5x^{5} \) is negative).

So, \( y \) can be any real number except \( 0 \), which means the range is \( (-\infty, 0)\cup(0, \infty) \).

Answer:

Domain: \( (-\infty, 0)\cup(0, \infty) \)
Range: \( (-\infty, 0)\cup(0, \infty) \) (corresponding to the option "Domain: \( (-\infty, 0)\cup(0, \infty) \), Range: \( (-\infty, 0)\cup(0, \infty) \)")