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
f) discuss the concavity
g) find any inflection points

Please show me how u got this!

I will assume you meant:

f(x) = (x^2 + 1)/(x^2 - 9)
= (x^2 + 1)/((x-3)(x+3))

a) for y intercepts , let x = 0
then y = 1/-9 = - 1/9
for x intercepts, let y = 0
(x^2 + 1)/(x^2 - 9) = 0
then x^2 + 1 = 0
x^2 = -1 , which has no real solutions
so there are no x-intercepts

b) using the quotient rule
dy/dx = ( (x^2-9)(2x) - (x^2+1)(2x))/(x^-9)^2
= -20x/(x^2 - 9)^2

c) --- critical values are found in many of the other parts.

d) for local extrema, dy/dx = 0
-20x/(x^2 - 9)^2 = 0
-20x = 0
x = 0 , then f(0) = -1/9

time to look at the graph using Wolfram, ( my other answers are confirmed here)
http://www.wolframalpha.com/input/?i=y+%3D+%28x%5E2+%2B+1%29%2F%28x%5E2+-+9%29

there is a local maximum at -1/9 , when x = 0

e) y '' = ((x^2 - 9)^2 (-20) + 20x(2)(x^2 - 9)(2x) )/(x^2 - 9)^4
= 60(x^2 + 3)/(x^2 - 9)^3

f) concave up or concave down depends on y''
if y'' is > 0 for a given x, then the curve is concave up at that value
if y'' is < 0 for .... concave down
While the numerator of y’’ is always positive,
the denominator of y’’ could be + or –
for -3 < x < 3 , the value of (x^2 – 9)^3 is negative, all all values yield a positive result.
So the curve is concave down for -3 < x < 3
For all other values it is concave up
Of course it is undefined for x = ± 3, giving us vertical asymptotes at
X = 3 and x = -3

g)
points of inflection occur when y'' = 0
60(x^2 + 3)/(x^2 - 9)^3 = 0
60(x^2 + 3) = 0
x^2 + 3 = 0
x^2 = -3
No solution, thus no point of inflection

a) To find the y-intercept, set x = 0 and solve for y:

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

So the y-intercept is -1/9.

To find the x-intercepts, set y = 0 and solve for x:

0 = x^2 + 1 / x^2 - 9

Multiplying both sides by (x^2 - 9) will eliminate the denominator:

0 = x^4 - 9x^2 + x^2 - 9

Combining like terms:

0 = x^4 - 8x^2 - 9

This is a quadratic equation in terms of x^2. We can factor it using the difference of squares:

0 = (x^2 - 9)(x^2 + 1)

Setting each factor equal to zero:

x^2 - 9 = 0 => (x + 3)(x - 3) = 0
x + 3 = 0 => x = -3
x - 3 = 0 => x = 3

x^2 + 1 = 0 => no real solutions

Therefore, the x-intercepts are x = -3 and x = 3.

b) To find the first derivative, we take the derivative of the function with respect to x:

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

Using the quotient rule, the derivative is given by:

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

Simplifying this expression will give us the first derivative.

c) To find the critical values, we set the first derivative equal to zero:

f'(x) = 0

We solve this equation to find the values of x that make the first derivative zero. These values will correspond to any potential critical points.

d) To find the local extrema, we evaluate the function at the critical values obtained in step c). The maximum and minimum values, if they exist, will be the local extrema.

e) To find the second derivative, we take the derivative of the first derivative:

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

Using the chain rule and product rule, we differentiate the expression for the first derivative. Simplifying this will give us the second derivative.

f) To discuss the concavity, we can use the second derivative test. For values of x where the second derivative is positive, the graph is concave up. For values where the second derivative is negative, the graph is concave down.

g) To find any inflection points, we set the second derivative equal to zero:

f''(x) = 0

Solving this equation will give us the values of x where an inflection point may occur.

a) To find the y-intercept, substitute x = 0 into the function:

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

Therefore, the y-intercept is (0, -1/9).

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

0 = x^2 + 1 / x^2 - 9

Multiply both sides by (x^2 - 9):

0 = (x^2 + 1)(x^2 - 9)

Using the difference of squares, we can rewrite it as:

0 = (x^2 + 1)(x + 3)(x - 3)

Setting each factor equal to zero:

x^2 + 1 = 0
x^2 = -1 (no real solutions)

x + 3 = 0
x = -3

x - 3 = 0
x = 3

Therefore, the x-intercepts are (-3, 0) and (3, 0).

b) To find the first derivative, differentiate the function with respect to x:

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

Simplifying:

f'(x) = (2x^3 - 18x - 2x^3) / (x^2 - 9)^2
= -18x / (x^2 - 9)^2

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

-18x / (x^2 - 9)^2 = 0

This equation is satisfied when x = 0. So x = 0 is a critical value.

d) To find the local extrema, we need to check the sign of the first derivative around each critical point.

For x < -3, choose x = -4:
-18(-4) / ((-4)^2 - 9)^2 = 72 / 25 > 0

For -3 < x < 0, choose x = -1:
-18(-1) / ((-1)^2 - 9)^2 = -18 / 64 < 0

For 0 < x < 3, choose x = 1:
-18(1) / (1^2 - 9)^2 = 18 / 64 > 0

For x > 3, choose x = 4:
-18(4) / (4^2 - 9)^2 = -72 / 25 < 0

From the analysis above, we can conclude that the function has a local minimum at (-3, 0) and a local maximum at (3, 0).

e) To find the second derivative, differentiate the first derivative with respect to x:

f''(x) = (-18(x^2 - 9)^2 - 2(-18x)(2x)) / (x^2 - 9)^4
= (-18(x^2 - 9)^2 + 36x^2) / (x^2 - 9)^4

f''(x) can be simplified further, but since it is not asked in the question, I will refrain from doing so.

f) To discuss the concavity, we need to analyze the sign of the second derivative.

For x < -3, choose x = -4:
(-18((-4)^2 - 9)^2 + 36(-4)^2) / ((-4)^2 - 9)^4 = -19004 / 625 < 0

For -3 < x < 0, choose x = -1:
(-18((-1)^2 - 9)^2 + 36(-1)^2) / ((-1)^2 - 9)^4 = -8560 / 6561 < 0

For 0 < x < 3, choose x = 1:
(-18((1)^2 - 9)^2 + 36(1)^2) / ((1)^2 - 9)^4 = -8560 / 6561 < 0

For x > 3, choose x = 4:
(-18((4)^2 - 9)^2 + 36(4)^2) / ((4)^2 - 9)^4 = -19004 / 625 < 0

From the analysis above, we can conclude that the function is concave down everywhere.

g) To find inflection points, set the second derivative equal to zero and solve for x. However, in this case, the second derivative is a fraction, and it can have no real solutions. Therefore, there are no inflection points for this function.