Asymptote at x=4 indicates a factor of (x-4) in the denominator.
Let's try f(x)=1/(x-4)
f'(x)=-1/(x-4)²
so f'(x)<0 for all x except 4.
Let's try f(x)=-1/(x-4)
f'(x)=1/(x-4)² >0 for all x.
Check f"(x)=2/(x-4)³
f"(x)>0 ∀x<4.
So all conditions are satisfied.
Which of the following could represent a function having the given properties?
1)increasing slope for x<4
2) f'(x)>0 for all x(x cannot =4)
3)asymptote at x=4.
Again I know we cannot graph but how can I figure this out?
1 answer