Need help on this curve-sketching problem.

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SUMMARY

The discussion focuses on curve sketching for the function f(t) = 3t^4 + 4t^3. Key findings include the domain being all real numbers, x-intercepts at (0,0) and (-4/3), and a y-intercept at (0,0). The function has no asymptotes, with end behavior pointing up at both ends. The first derivative is f'(t) = 12t^3 + 12t^2, indicating decreasing behavior on (-∞, -1) and increasing on (-1, +∞), with a relative minimum at (-1, -1). The second derivative is f''(t) = 36t^2 + 24t, showing concavity changes at (-2/3, 0) and points of inflection at (0,0) and (-2/3, -16/27).

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Would someone please help me on this problem? I'm sorry if it is hard to read. I'm new to this forum and haven't had the time to read over the latex system guide. I just need to make sure the information that I got is correct. Thank you.

Sketch the graph of the function, using the curve-sketching guide.

Function: f(t)=3(t)^(4) + 4(t)^(3)

From this function, I derived this information (Please check):
Domain: all real #s
x-int: (0,0)
y-int: (0,0)
Asymptote: none
End Behavior: points up at both ends.
First Derivative: f'(t)=12(t)^(3) + 12(t)^(2)
Decreasing: (-infinity,-1)
Increasing: (-1,+infinity)
Relative Minima: (-1,-1)
This is where I think I messed up!
Second Derivative: f''(t)=36(t)^(2) + 24(t)
Concave Up: (0,+infinity), (-infinity,-2/3)
Concave Down: (-2/3,0)
Points of Inflection: (0,0), (-2/3,-16/27)
 
Last edited:
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Perfect. Got the same thing.

EDIT: You missed one x-intercept at x = -4/3. Besides that, looks good.
 
Jameson said:
Perfect. Got the same thing.

EDIT: You missed one x-intercept at x = -4/3. Besides that, looks good.
Thanks a lot!
 

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