# antiderivatives

• Nov 6th 2012, 12:53 PM
pnfuller
antiderivatives
find a function f such that f'(x)=x3 and the line x+y=0 is tangent to the graph of f
• Nov 6th 2012, 02:10 PM
skeeter
Re: antiderivatives
$f'(x) = x^3 \implies f(x) = \frac{x^4}{4} + C$

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

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

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

finally ...

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

... solve for C to finish it.