Calculus Graph wit the conditions listed below

Question

f(2)=3

f ' (x)>0 if x does not equal 2

f " (x) >0 if x<2

f " (x)<0 if x>2

lim f(x)=8

as x approaches oo

lim f(x)=0.

as x approaches -oo

Answer 1

Whenever asked to graph, I always graph what I know first. There is a point at (2,3). But the derivatives do not exist at x=2. So there is either an asymptote at x=2 or a hole at x=2 or a sharp turn at x=2 or a vertical slope at x=2. Because there exists a point at x=2, I would guess that this is either a sharp turn or a vertical slope.

The limits at infinities give us the equations of the horizontal asymptotes. The one on the right is at y=8, the one on the left is at y=0.

first deriv is pos everywhere except for 2. So this function is increasing everywhere (except for x=2).

for x<2, the 2nd deriv is pos so is concave up. for x>2, 2nd deriv is neg so f(x) is concave down.

1st deriv pos & 2nd deriv pos looks like the right side of a "U" -- it is increasing and concave up. Now flip the "U" upside down, and use the left side - this is the 1st deriv pos and 2nd deriv neg -- increasing and concave down.

Putting it all together, you should have a horizontal asymptote drawn at y = 8, from 0 to +inf

And another horizontal asymptote from -inf to 0 at y=0, and you should have a point at (2,3).

The derivative information suggest something like an elongated "S" = top hugs asymptote at y=8 as it heads out to +inf. Bottom hugs asymptote at y=0 ans it goes back to -inf. The each make their way toward (2,3) but become vertical as they pass through (2,3).

I wish I could draw this or add a picture - it would not let me record a video.

__________

/

|

__________/