Answer:
a)[tex]X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)[/tex]
b)[tex]Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0 )[/tex]
Step-by-step explanation:
From the question we are told that
The Function
[tex]f(x)=1+\frac{1}{x} +\frac{5}{x^2} +\frac{1}{x^3}[/tex]
Generally the differentiation of function f(x) is mathematically solved as
[tex]f(x)=1+\frac{1}{x} +\frac{5}{x^2} +\frac{1}{x^3}[/tex]
[tex]f(x)=\frac{x^3+x^2+5x+1}{x^2}[/tex]
Therefore
[tex]f'(x)=\frac{x^2+10x+3}{x^4}[/tex]
Generally critical point is given as
[tex]f'(x)=0[/tex]
[tex]\frac{x^2+10x+3}{x^4}=0[/tex]
[tex]x=-5 \pm\sqrt{22}[/tex]
Generally the maximum and minimum x value for critical point is mathematically solved as
[tex]f'(-5 \pm\sqrt{22})[/tex]
Where
Maximum value of x
[tex]f'(-5 +\sqrt{22})<0[/tex]
Minimum value of x
[tex]f'(-5 +\sqrt{22})<0[/tex]
Therefore interval of increase is mathematically given by
[tex]f'(-5 -\sqrt{22}),f'(-5 +\sqrt{22})[/tex]
[tex]f(x)<0,-\infty<x<(f'(-5 -\sqrt{22})) ,(f'(-5 +\sqrt{22}))<x<0,0<x< \infty[/tex]
Therefore interval of decrease is mathematically given by
[tex](-\infty,-5 -\sqrt{22}),f'(-5 +\sqrt{22},0),(0,\infty)[/tex]
Generally the second differentiation of function f(x) is mathematically solved as
[tex]f''(x)=\frac{2(x^2+15x+6)}{x^5}[/tex]
Generally the point of inflection is mathematically solved as
[tex]f''(x)=0[/tex]
[tex]x^2+15x+6=0[/tex]
Therefore inflection points is given as
[tex]x=\frac{1}{2} (-15 \pm \sqrt{201}[/tex]
[tex]f''(x)>0,\frac{1}{2}(-15-\sqrt{201}) <x<\frac{1}{2}(-15-\sqrt{201}) <x<0[/tex]
a)Generally the concave upward interval X is mathematically given as
[tex]X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)[/tex]
[tex]f''(x)<0,-\infty<x<\frac{1}{2}(-15-\sqrt{201}) , \frac{1}{2}(-15-\sqrt{201}) <x<0[/tex]
b)Generally the concave downward interval Y is mathematically given as
[tex]Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0 )[/tex]
