Question

The function f(x)=(1)/(x^2−25) is not defined at these x values:

a) x=5 and x=−5
b) x=0
c) x=5
d) x=±25

Answer 1

a)
`\( x = 5 \) and \( x = -5 \)`

The function
`\( f(x) = (1)/(x^2 - 25) \)`is not defined at
`\( x = 5 \) and \( x = -5 \)`because these values result in a zero denominator, leading to division by zero, which is undefined in mathematics.

Step-by-step explanation:

The given function is
`\( f(x) = (1)/(x^2 - 25) \)`. To determine where the function is not defined, we need to identify values of
`\( x \)`that make the denominator equal to zero. In this case, the denominator is
`\( x^2 - 25 \)`. Setting this expression equal to zero, we find:

`\[ x^2 - 25 = 0 \]`

Factoring the quadratic expression, we get \
`( (x + 5)(x - 5) = 0 \)`. This equation has solutions
`\( x = 5 \) and \( x = -5 \).` Therefore, the function is not defined at
`\( x = 5 \) and \( x = -5 \).`

In summary, the function
`\( f(x) = (1)/(x^2 - 25) \)` is not defined at
`\( x = 5 \) and \( x = -5 \)`. This is because these values make the denominator zero, leading to division by zero, which is undefined in mathematics.