For f(x) = 1 + 1/x + 9/x^2 + 1/x^3 Use calculus to find the intervals of increas

Question

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

Use calculus to find the intervals of increase and decrease and the intervals of concavity. (Enter your answers in interval notation. Do not round your answers.)

1. Find the interval of increase. (PLEASE ANSWER IN INTERVAL NOTATION)

2. Find the interval of decrease. (PLEASE ANSWER IN INTERVAL NOTATION)

3. Find the interval where the function is concave up. (PLEASE ANSWER IN INTERVAL NOTATION)

4. Find the interval where the function is concave down. (PLEASE ANSWER IN INTERVAL NOTATION)

Explanation / Answer

f'(x)=-1/x^2-18/x^3-3/x^4=-1/x^4(x^2+18x+3)=-1/x^4 (x-(-9+sqrt(78))) (x-(-9-sqrt(78)))

so

a) inc in x (-9-sqrt(78) , -9+sqrt(78))

b) dec in x>-9+sqrt(78) and x<-9-sqrt(78)

f''(x)=-2/x^3-54/x^4-12/x^5 = -2/x^5(x^2+27x+6)=-2/x^5 (x-(-27/2-sqrt(705)/2)) (x-(-27/2+sqrt(705)/2))

concave down in x>0; -27/2-sqrt(705)/2<x<-27/2+sqrt(705)/2

and concave up else where

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