Please answer the following questions about the function Please answer the follo

Question

Please answer the following questions about the function

Please answer the following questions about the function x = y =- Vertical asymptotes f. Horizontal asymptotes x = (c) Find any horizontal and vertical asymptotes of f is concave up, concave down, and has inflection points. Concave up on the interval Concave down on the interval Inflection points x = (b) Find where x = Local minima x = Increasing on the interval Decreasing on the interval Local maxima f , where it is increasing and decreasing, and its local extrema. Critical numbers y -values, enter either a number, a list of numbers separated by commas, or None if there aren't any solutions. Use interval notation if you are asked to find an interval or union of intervals, and enter { } if the interval is empty. (a) Find the critical numbers of x - or f(x) = (x^2 + 10)(9 - x^2). Instructions: If you are asked to find

Explanation / Answer

f(x) = (x^2 + 10)(9 - x^2)

f(x) = -x^4 + 9x^2 - 10x^2 + 90

f(x) = -x^4 - x^2 + 90

a)

f'(x) = -4x^3 - 2x = 0

4x^3 + 2x = 0

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

2x = 0 , 2x^2 + 1 = 0

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

f''(x) = -12x^2 - 2
f''(0) = -2 ---> negative
So, x = 0 corresponds to LOCAL MAXIMUM

When x = 0, y = (x^2 + 10)(9 - x^2) becomes y = (0 + 10)(9 - 0) ---> y = 90
No local minimum value exists

Critical value, x = 0

So, the regions become (-inf , 0) and (0 ,inf)

Region 1 : (-inf , 0)
Testvalue = -1
f'(x) = -4x^3 - 2x
f'(-1) = 4 + 2
f'(-1) = 6 --> positive
So, increasing over (-inf , 0)

Region 2 : (0 , inf)
Testvalue = 1
f'(x) = -4x^3 - 2x
f'(1) = -4 - 2
f'(1) = -6 --> negative
So, decreasing over (0 , inf)

So, here are the answers :

Critical numbers, x = 0 ---> ANSWER
Increasing : (-inf , 0) --> ANSWER
Decreasing : (0 , inf) --> ANSWER
Local maxima, x = 0 --> ANSWER
Local minima, x = None --> ANSWER

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b)

f''(x) = -12x^2 - 2 = 0

12x^2 = -2

x^2 = -2/12

x^2 = -1/6

No solution

So, no inflection points

So, region is (-inf , inf)
Testvalue = 0
f''(x) = -12x^2 - 2
f''(0) = -2 --> negative
So, concave down everywhere

Concave up nowhere

Concave up : None ---> ANSWER
Concave down : (-inf , inf) --> ANSWER
Inflection points, x = NONE --> ANSWER

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c)

f(x) = (x^2 + 10)(9 - x^2)

This is a polynomial function, which NEVER has any asympttoes

So, horizontal asymptotes : y = NONE --> ANSWER
Vertical asymptotes, x = NONE --> ANSWER

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