Essay Lab | Math Shack | Videos
TABLE OF CONTENTS
Draw a graph of
• Find dots.
f is zero only when x = 0. We find an interesting new twist here, because f has asymptotes:
there's a vertical asymptote at x = 4 and a horizontal asymptote at y = 12.
The derivative of f is
This is only zero or undefined when x = 4. Since f is undefined when x = 4, we don't find a critical point here, so f has no critical points.
The second derivative of f is
Again, the only place f " is zero or undefined is at x = 4, where f is undefined. Therefore f has no inflection points either.
When we go to graph the dots, we also put on the asymptotes:
• Find shapes.
Make the numberlines:
The function is negative when 0 < x < 4 (since then 12x is negative and ( x - 4 ) is positive), and positive everywhere else:
The derivative f '(x) = -48( x - 4 )2 is always negative, since ( x – 4 )2 is always positive.
The second derivative is negative when ( x – 4 ) is negative and positive when ( x – 4 ) is positive:
By looking at f ' and f " we can figure out the shapes of f :
• Play connect-the-dots.
We connect the dots to find the final graph, remembering to use the asymptotes: