Solving Function Given Tangent Line: x+y=0

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Homework Help Overview

The problem involves finding a function \( f \) such that its derivative \( f'(x) = x^3 \) and the line \( x + y = 0 \) is tangent to the graph of \( f \). The context is centered around calculus, specifically dealing with derivatives and tangent lines.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the antiderivative of \( f' \) and the implications of the tangent line's slope. There is an exploration of how the slope of the tangent relates to the derivative and the function itself. Questions arise regarding the specific point of tangency and the corresponding values of \( f \).

Discussion Status

The discussion is ongoing, with participants clarifying the relationship between the slope of the tangent line and the derivative of the function. Some guidance has been offered regarding the interpretation of the tangent line's slope and its implications for the function.

Contextual Notes

Participants are navigating the relationship between the function, its derivative, and the tangent line, with some uncertainty about how to proceed from the derived antiderivative and the conditions set by the tangent line.

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Homework Statement


Find a function f such that f'(x)=x^3 and the line x+y=0 is tangent to the graph of f.



Homework Equations


Need antiderivative


The Attempt at a Solution


I found the antiderivative, aka the function, to be (x^4)/4 + C
Since the tangent like is x+y=0, --> y=-x, so the slope is -1
If the slope is -1, that means the derivative, x^3 also equals -1
Plugging in f(-1) gives 1/4 + C

Now I don't know what to do with that.
 
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It is saying that the tangent to f is -1, not f itself.

Casey
 
Saladsamurai said:
It is saying that the tangent to f is -1, not f itself.

Casey

Well wouldn't it still be the same slope anyhow?
 
Where (what x) is the slope of x4/4+ C equal to -1? If x+ y= 0 is tangent to y= x4/4+ C they must have the same y value there.
 

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