Stationary points are points at which f'(x)=0 AND f"(x)=0.
Vertical asymptote is where the denominator becomes zero.
Horizontal asymptote is the limit when x->∞ or x->-∞, if the limit is defined.
Given [(x-5)/(x-3)]^2 . Find any stationary points and any points of inflection. Also find any horizontal and
vertical asymptotes.
3 answers
could you please solve the problem completely .. i am still confused.
Thanks.
Thanks.
Denominator becomes zero when x=3, therefore vertical asymptote is a x=3.
The limit of f(x)=[(x-5)/(x-3)]^2 as x->±∞ is 1, so the horizontal asymptote is at y=1.
Since f'(5)=0 and f"(6)=0, there are no stationary points, just a minimum at x=5.
There is a point of inflection at x=6 (where f"(6)=0).
You will need to take the time to calculate f'(x) and f"(x) from f(x), and hence sketch the graph of f(x).
A sketch of the graph can be seen here:
http://img718.imageshack.us/img718/4296/1292787276.png
The limit of f(x)=[(x-5)/(x-3)]^2 as x->±∞ is 1, so the horizontal asymptote is at y=1.
Since f'(5)=0 and f"(6)=0, there are no stationary points, just a minimum at x=5.
There is a point of inflection at x=6 (where f"(6)=0).
You will need to take the time to calculate f'(x) and f"(x) from f(x), and hence sketch the graph of f(x).
A sketch of the graph can be seen here:
http://img718.imageshack.us/img718/4296/1292787276.png
1 answer