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

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- Nov 6th 2012, 12:53 PMpnfullerantiderivatives
find a function f such that f'(x)=x

^{3}and the line x+y=0 is tangent to the graph of f - Nov 6th 2012, 02:10 PMskeeterRe: antiderivatives
$\displaystyle f'(x) = x^3 \implies f(x) = \frac{x^4}{4} + C$

$\displaystyle x+y = 0 \implies y = -x$ ... slope of the tangent line is $\displaystyle m = -1$

at the point of tangency $\displaystyle (x_0,y_0)$ ...

$\displaystyle f'(x_0) = x_0^3 = -1 \implies x_0 = -1$

finally ...

$\displaystyle y_0 = -x_0 = -(-1) = \frac{(-1)^4}{4} + C$

... solve for C to finish it.