(a) Find the domain. (b) Find any x and y intercepts. (c) Is the function even o

Question

(a) Find the domain.

(b) Find any x and y intercepts.

(c) Is the function even or odd? Is it periodic?

(d) Find any/all vertical and horizontal asymptotes.

(e) Find the first derivative and list all intervals of increase and decrease.

(f) List any/all local max and min points.

(g) Find the second derivative and list all intervals where f is concave up and concave down. List any/all inflection points.

(h) Sketch the graph. Label all intercepts, asymptotes, local max and min points and points of inflection.

h(x)=x^2-4/9-x^2 (a) Find the domain. (b) Find any x and y intercepts. (c) Is the function even or odd? Is it periodic? (d) Find any/all vertical and horizontal asymptotes. (e) Find the first derivative and list all intervals of increase and decrease. (f) List any/all local max and min points. (g) Find the second derivative and list all intervals where f is concave up and concave down. List any/all inflection points. (h) Sketch the graph. Label all intercepts, asymptotes, local max and min points and points of inflection.

Explanation / Answer

h(x) = (x^2 - 4) / (9 - x^2)

a)

h(x) = (x + 2)(x - 2) / (3 - x)(3 + x) ---> factored form

Nothing cancels between numerator and denominator

So, to get domain, equate denominator to 0
(3 - x)(3 + x) = 0
x = -3 and 3

So, domain : (x cannot be -3 or 3}

(-inf , -3) U (-3 , 3) U (3 , inf) -----> ANSWER

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B)

x-intercepts :
Equate y with 0 and solve for x
(x + 2)(x - 2) / (3 - x)(3 + x) = 0
(x + 2)(x - 2) = 0
x = 2 and -2 ---> ANSWER

y-intercept :
Plug in x with 0 and find y
(0 + 2)(0 - 2) / (3 - 0)(3 + 0)
y = -4 / 9 ---> ANSWER

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C)

h(x) = (x^2 - 4) / (9 - x^2)

First we find h(-x)

h(-x) = ((-x)^2 - 4) / (9 - (-x)^2)

h(-x) = (x^2 - 4) / (9 - x^2) = h(x)

Since h(-x) = h(x), the function is EVEN ---> ANSWER

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D)

VA :
Domain was {X cannot be -3 and 3}
So, VA: x = -3 and x = 3 ---> ANSWER for VA

HA :
Degree of numerator = 2
Degree of denominator = 2
Since degrees are equal, the HA is just the ratio of the leading terms...
HA : y = 1x^2 / -1x^2 ---> y = -1 ---> ANSWER for HA

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E)

Inc.dec :

h(x) = (x^2 - 4) / (9 - x^2)

Deriving using quotient rule :

u = x^2 - 4 , v = 9 - x^2
u' = 2x , v' = -2x

(u'v - uv') / v^2

(18x - 2x^3 + 2x^3) / (9 - x^2)^2

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

18x = 0

x = 0 ---> this is the critical value

And VA was -3 and 3

So, this splits the number line into (-inf , -3) , (-3 , 0) , (0 , 3) and (3 , inf)

Region 1 : (-inf , -3)
Testvalue = -4
h'(x) = 18x / (9 - x^2)^2
h'(-4) --> negative
So, decreasing

Region 2 : (-3 , 0)
Decreasing

Region 3 : (0 , 3)
Increasing

Region 4 : (3 , inf)
Increasing

So, decreasing : (-inf , -3) U (-3 , 0) --> ANSWER
increasing : (0 , 3) U (3 , inf) --> ANSWER

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F)

The only critical value was x = 0, found in part E above
Now, at x = 0, decrasing changs to increasing
So, x = 0 ---> point of local minimum

h( x) = (x^2 - 4) / (9 - x^2)

So, h(0) = -4/9

So, the local minimum is (0 , -4/9) ----> ANSWER
No local maximums ---> ANSWER

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G)

h'(x) = 18x / (9 - x^2)^2

h''(x) = 54(x^2 + 3) / (9 - x^2)^3 = 0

54(x^2 + 3) = 0

x^2 + 3 = 0 ---> this has no solution

So, no inflection points

h''(x) = 54(x^2 + 3) / (9 - x^2)^3

Region 1 : (-inf , -3)
Testvalue = -4
h''(-4) = negative
So, concave down

Region 2 : (-3 , 3)
Testvalue = 0
h''(0) = positive
So, concave up

Region 3 : (3 , inf)
Tetvalue = 4
Concave down

So, concave down : (-inf , -3) U (3 , inf) --> ANSWER
Concave up : (-3 , 3) ---> ANSWER

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