Help with Curve Sketching - Solve in 65 Characters

[Slimsta]

Curve Sketching help!

Homework Statement

http://img21.imageshack.us/img21/7660/54035638.jpg

Homework Equations

The Attempt at a Solution

For part E, I've been sitting on it for the last 2 hours trying to figure out why this is wrong..
i came up with a solution but it doesn't seem to be right..
if solving for f'(x) i get x=6.279
so f'(x) > 0 ==> x<6.279
f'(x) < 0 ==> x>6.279
decreasing (-inf, 6.279)U(6.279, inf) and increasing on (6.279, 0) ? I am not sure about the increasing part..
but when graphing it on my calc, it look like 1.521 would work better..

 

[Mark44]

I drew a very rough sketch showing the two asymptotes and the zero at x = .39. For all x < .39, f(x) < 0, right? Also, f(0) \approx -.02 from your work.

As x --> -\infty, f(x) --> 0 and is negative. So somewhere way out to the left, the function values are less negative (larger) than the one at x = 0. This means that, going left to right, the graph of f is decreasing for a while, levels off, and then heads up to (0, -.02) and continues on up. The upshot is that the tangent line has to be horizontal at some negative value of x (not anywhere near 6.3). Check your differentation work.

 

[Slimsta]

so it would be,
decreasing on (-inf, 0)U(0, inf) and increasing on (0, -0.02)
right?

 

[Mark44]

No. The graph is decreasing on (-inf, A) U (1.521, inf), and increasing on (A, 1.521). You have to find the number A, and it is less than zero.

 

[Slimsta]

No. The graph is decreasing on (-inf, A) U (1.521, inf), and increasing on (A, 1.521). You have to find the number A, and it is less than zero.

the only number that is less than 0 is -0.016858 from this question.. and that doesn't work :(

 

[Mark44]

In your OP you said that g'(6.279) = 0. That is wrong, as I said in post #2. What do you have for g'(x)?

 

[Slimsta]

In your OP you said that g'(6.279) = 0. That is wrong, as I said in post #2. What do you have for g'(x)?

for g' i get
g'(x) = (-100x + 627) / (10x-15.21)²

if g'(x) = 0 then x=6.27
i tried doing the derivative 5 times already and i got the same result

 

[erok81]

At first glance that derivative isn't correct. Shouldn't the bottom be ^4 not ^2?

 

[Slimsta]

At first glance that derivative isn't correct. Shouldn't the bottom be ^4 not ^2?

it wouldn't matter anyways..
i need the top to be equal to 0.
-100x + 627 = 0
x=6.27

 

[erok81]

I don't think the top is correct either.

How are you arriving at your derivative?

 

[Slimsta]

(f'g - g'f) / g^2

so f(x) = 10x-3.9
and g(x) = (10x-15.21)^2

[10(10x-15.21)^2 ] - [2(10x-15.21)*10*(10x-3.9)] / [(10x-15.21)^2]^2
==> [10(10x-15.21) - 20(10x-3.9) ] / (10x-15.21)^3
==> (100x - 152.1 - 200x + 780) / (10x-15.21)^3
==> (-100x + 627) / (10x-15.21)^3

 

[Mark44]

(f'g - g'f) / g^2

so f(x) = 10x-3.9
and g(x) = (10x-15.21)^2

[10(10x-15.21)^2 ] - [2(10x-15.21)*10*(10x-3.9)] / [(10x-15.21)^2]^2
==> [10(10x-15.21) - 20(10x-3.9) ] / (10x-15.21)^3
==> (100x - 152.1 - 200x + 780) / (10x-15.21)^3
==> (-100x + 627) / (10x-15.21)^3

Error in the 3rd line: the 780 should be 78, since you have 20*3.9, not 20*39.

So g'(x) = (-100x - 74.1)/(10x - 15.21)^3

If you recall, I've been saying that the derivative is zero for some negative value of x.

 

[Slimsta]

Error in the 3rd line: the 780 should be 78, since you have 20*3.9, not 20*39.

So g'(x) = (-100x - 74.1)/(10x - 15.21)^3

If you recall, I've been saying that the derivative is zero for some negative value of x.

oh it my mistake again.. haven't paid attention to it :/
g'(x) = 0 when x=-0.741

so now i know that there is a local min at x=-0.741

for part G, it will concave up (-inf, 1.521) and concave down (-0.741, 1.521)U(1.521, inf)
the IP is x=-1.872

right?

 

[Mark44]

oh it my mistake again.. haven't paid attention to it :/
g'(x) = 0 when x=-0.741

Right.

so now i know that there is a local min at x=-0.741

Right.

for part G, it will concave up (-inf, 1.521) and concave down (-0.741, 1.521)U(1.521, inf)
the IP is x=-1.872

right?

The inflection point seems about right - I haven't worked that out, but where you have it seems to be the right place. Your concavity descriptions are all wrong, though. The whole idea of an inflection point is that it's where the concavity changes. To the left of the inflection point, the graph is concave down. Between the inflection point and 1.521, the graph is concave up, and for x > 1.521, the graph is also concave up.

 

[Slimsta]

right right right! i got it before i read the message.. i wish i saw it earlier to save time though :b

thank you so much once more!