Given the function: f(x) = x^2 + 1 / x^2 - 9

a)find y and x intercepts
b) find the first derivative
c) find any critical values
d) find any local(relative) extrema
e) find second derivative

Please show me how u got this!!

a) To find the y-intercept, we set x = 0 in the function. So, we get:

f(0) = (0^2 + 1) / (0^2 - 9)
= 1 / (-9)
= -1/9

Thus, the y-intercept is -1/9.

To find the x-intercepts, we set f(x) = 0 and solve for x. So, we have:

x^2 + 1 / x^2 - 9 = 0
x^2 + 1 = 9
x^2 = 8
x = ±√8

Therefore, the x-intercepts are x = √8 and x = -√8.

b) To find the first derivative, we differentiate the function with respect to x. So, we have:

f'(x) = d/dx [(x^2 + 1) / (x^2 - 9)]
= [(2x)(x^2 - 9) - (x^2 + 1)(2x)] / (x^2 - 9)^2

Simplifying this expression gives us the first derivative of f(x).

c) To find the critical values, we set the first derivative equal to zero and solve for x. So, we have:

f'(x) = 0
[(2x)(x^2 - 9) - (x^2 + 1)(2x)] / (x^2 - 9)^2 = 0

Solve this equation to find the critical values of x.

d) To find the local (relative) extrema, we evaluate the first derivative at the critical values we found in step c. If the first derivative changes sign from positive to negative or from negative to positive, we have a local extremum at that point.

e) To find the second derivative, we differentiate the first derivative with respect to x. So, we have:

f''(x) = d/dx [f'(x)]

Simplify this expression to find the second derivative of f(x).