Question

consider the function and then use calculus to answer the questions that follow 1 1/x 5/x^2 1/x^3 (a) Find the interval(s) where the function is concave up. (Enter your answer in interval notation,do not round). (b) Find the interval(s) where the function is concave down. (Enter your answer in intervalnotation, do not round).

Answer 1

a)
`X=((-15-√(201),(-15+√(201)),(0,\infty)`

b)
`Y=(\infty,(1)/(2)(-15-√(201) ) ),((1)/(2)()-15+√(201)),0 )`

Explanation:

From the question we are told that

The Function

`f(x)=1+(1)/(x) +(5)/(x^2) +(1)/(x^3)`

Generally the differentiation of function f(x) is mathematically solved as

`f(x)=1+(1)/(x) +(5)/(x^2) +(1)/(x^3)`

`f(x)=(x^3+x^2+5x+1)/(x^2)`

Therefore

`f'(x)=(x^2+10x+3)/(x^4)`

Generally critical point is given as

`f'(x)=0`

`(x^2+10x+3)/(x^4)=0`

`x=-5 \pm√(22)`

Generally the maximum and minimum x value for critical point is mathematically solved as

`f'(-5 \pm√(22))`

Where

Maximum value of x

`f'(-5 +√(22))<0`

Minimum value of x

`f'(-5 +√(22))<0`

Therefore interval of increase is mathematically given by

`f'(-5 -√(22)),f'(-5 +√(22))`

`f(x)<0,-\infty<x<(f'(-5 -√(22))) ,(f'(-5 +√(22)))<x<0,0<x< \infty`

Therefore interval of decrease is mathematically given by

`(-\infty,-5 -√(22)),f'(-5 +√(22),0),(0,\infty)`

Generally the second differentiation of function f(x) is mathematically solved as

`f''(x)=(2(x^2+15x+6))/(x^5)`

Generally the point of inflection is mathematically solved as

`f''(x)=0`

`x^2+15x+6=0`

Therefore inflection points is given as

`x=(1)/(2) (-15 \pm √(201)`

`f''(x)>0,(1)/(2)(-15-√(201)) <x<(1)/(2)(-15-√(201)) <x<0`

a)Generally the concave upward interval X is mathematically given as

`X=((-15-√(201),(-15+√(201)),(0,\infty)`

`f''(x)<0,-\infty<x<(1)/(2)(-15-√(201)) , (1)/(2)(-15-√(201)) <x<0`

b)Generally the concave downward interval Y is mathematically given as

`Y=(\infty,(1)/(2)(-15-√(201) ) ),((1)/(2)()-15+√(201)),0 )`