Question

f(x)=1+7/x-4/x^2

where is the function increasing/decreasing?

Answer 1

find first derivative. Later look for intervals of x axis where this derivative is negative/positive, since that means that function is decreasing/increasing there

Answer 2

for the derivative i get \[\frac{ -7x+8 }{ x^3 }\] so x=8/7

Answer 3

there is a vertical asymptote at 1 and a horizontal at 0

Answer 4

\[f'(x)=-\frac{7}{x^{2}}+\frac{8}{x^{3}}\]
this is not defined at x=0. So you hae to study what happens for x>0 and x<0
1º
Let's start with x>0.
f'(x)=0 at 7/x^2=8/x^3 => x=8/7
for x>0 and x<8/7 f'(x)>0. so function increase
for x>0 and x>8/7 f'(x)<0. So function decrease
2º x<0
In this case f'(x)<0 for all x. So function decrease

Answer 5

got it?

Answer 6

so it would be
increasing:(0,8/7)
decreasing: (8/7, inf)?

Answer 7

no:
increasing:(0,8/7)
decreasing: (-inf,0) union (8/7, inf)

Answer 8

@guest1234