Thread: Asymptotes and Limits of Infinity

Directions: sketch the graph.

f(t) = 3t^4 - 4t^2 + 3

This is the first problem I've had like this. Since the polynominal equation isn't obvious, I think I take x to the limit of infinity:
f(t) 3t^4 - 4t^2 +3 = 3t^2 -4 +3/t^2 = 3t^2 = 0
lim approaches infinity +
t=0
f(0) = 3
extrema point at (0,3)

f'(t) = 12t^3 - 8t
f'(t) = 4t(3t^2 - 2)
f'(t) = 4t =0
f'(t) = t=0
f'(t) = 3t^2 -2 = 0
f'(t) = 3t^2 = 2
f'(t) = t^2 = 2/3
f'(t) = t = 2/3
f'(t) = 0.8164
f'(.8164) = 12(.8164)^3 - 8(.8164)
f'(.8164) = .6529 - 6.5312
f'(.8164) = -5.8783

critical points at (0,0) and (.8164, -5.8783)

f"(t) = 36t^2 - 8
f"(t) = 4(6t^2 - 2)
f"(t) = 4
f"(t) = 6t^2 - 2 = 0
f"(t) = 6t^2 = 2
f"(t) = t^2 = 1/3
f"(t) = t = 1/3
f"(t) = t =.577350269

f"(4) = 36(16) - 8
f"(4) = 568
(4,568)
f"(.577350268) = 36(.577350269)^2 - 8
f"(.577350269) = 12 - 8
f"(.577350269) = 4
(.577350269,4)

relative maximium (4,568)


Anything I am forgetting (or just plain wrong)?

Originally Posted by confusedagain

Directions: sketch the graph.

f(t) = 3t^4 - 4t^2 + 3

This is the first problem I've had like this. Since the polynominal equation isn't obvious, I think I take x to the limit of infinity:
f(t) 3t^4 - 4t^2 +3 = 3t^2 -4 +3/t^2 = 3t^2 = 0
lim approaches infinity +
t=0
f(0) = 3
extrema point at (0,3)

f'(t) = 12t^3 - 8t
f'(t) = 4t(3t^2 - 2)
f'(t) = 4t =0
f'(t) = t=0
f'(t) = 3t^2 -2 = 0
f'(t) = 3t^2 = 2
f'(t) = t^2 = 2/3
f'(t) = t = 2/3
f'(t) = 0.8164
f'(.8164) = 12(.8164)^3 - 8(.8164)
f'(.8164) = .6529 - 6.5312
f'(.8164) = -5.8783

critical points at (0,0) and (.8164, -5.8783)

f"(t) = 36t^2 - 8
f"(t) = 4(6t^2 - 2)
f"(t) = 4
f"(t) = 6t^2 - 2 = 0
f"(t) = 6t^2 = 2
f"(t) = t^2 = 1/3
f"(t) = t = 1/3
f"(t) = t =.577350269

f"(4) = 36(16) - 8
f"(4) = 568
(4,568)
f"(.577350268) = 36(.577350269)^2 - 8
f"(.577350269) = 12 - 8
f"(.577350269) = 4
(.577350269,4)

relative maximium (4,568)


Anything I am forgetting (or just plain wrong)?

Suppose I just show one way how to graph it. (I cannot follow very well your computions.]

To graph is to find the critical points, the asymptotes, concavity, inflection points, some other points, etc.

Critical points. Where f'(t) = 0:
f(t) = 3t^4 - 4t^2 + 3
f'(t) = 12t^3 -8t
0 = 12t^3 -8t
0 = 4t(3t^2 -2)
t = 0 or sqrt(2/3) or -sqrt(2/3)
t = 0, or 0.8165, or -0.8165 --------------at critical points.

f(0) = 3. So point (0,3)
f(0.8165) = 3(0.8165^4) -4(0.8165^2) +3 = 1.6667---so point (0.8165,1.6667)
f(-0.8165) = 3(-0.8165)^4 -4(-0.8165)^2 +3 = 1.6667---so point (-0.8165,1.6667)

Asymptotes: None. f(t) is polynomial in t, not in fraction form.

Concavity at critical points, or f''(t) at critical points:
f''(t) = 36t^2 -8

f''(0) = -8 ----negative, graph concaves facing downward.
f''(0.8165) = 36(0.8165^2) -8 = 16 ------positive, graph cocaves upward.
f''(-0.8165) = 36(-0.8165)^2 -8 = 16 ------positive, graph cocaves upward.

Plot those 3 critical points and sketch the concavities of the graph at those 3 points.
There is a change in concavity between (-0.8165,1.6667) and (0,3). Likewise, there is also a change in concavity between (0,3) and (0.8165,1.6667). So there must be two inflection points.

The inflection point:
f''(t) = 0
36t^2 -8 = 0
t = +,-sqrt(8/36) = +,-0.4714 ----***
f(+,-0.4714) = 3(+,-0.4714)^4 -4(+,-0.4714)^2 +3 = 2.2593
Meaning, the inflection points are (-0.4714,2.2593) and (0.4714,2.2593).

So now you have 5 points. Get at least two more. One to the left of (-0.8165,1.6667) and another to the right of (0.8165,1.6667).

Then connect those 7 points with a smooth curve. There is your graph of the function. [White-Westinghouse would love that graph.]