Domain of Function: Finding Valid Inputs for (x+4 5/7)²

Question

Find the positive and negative domains of the function below:

y=(x+457)2 y=\left(x+4\frac{5}{7}\right)^2

Step-by-Step Solution

To determine the positive and negative domains of the function y=(x+457)2 y = \left(x + 4\frac{5}{7}\right)^2 , we need to analyze the behavior of the expression inside the square.

The expression (x+457)2\left(x + 4\frac{5}{7}\right)^2 represents a perfect square. A perfect square is always greater than or equal to zero.

Consider the following observations:

  • The expression is zero when x=457 x = -4\frac{5}{7} . At this point, y=0 y = 0 .
  • For any other value of x x , x457 x \neq -4\frac{5}{7} , the expression (x+457)2\left(x + 4\frac{5}{7}\right)^2 is positive.

Given that (x+457)2 \left(x + 4\frac{5}{7}\right)^2 cannot be negative for any real x x , the function has no negative domain.

The positive domain of y y is all x x except when x=457 x = -4\frac{5}{7} . Hence, the positive domain is the set where x457 x \neq -4\frac{5}{7} .

In conclusion, the negative domain is none, and the positive domain is x457 x \neq -4\frac{5}{7} .

Therefore, the correct choice is:

x < 0 : none

x > 0 : x\ne-4\frac{5}{7}

Answer

x < 0 : none

x > 0 : x\ne-4\frac{5}{7}