Finding a function satisfying two tangents?

Question

I want to find a function $f$ that satisfies the following conditions:

(i) $f'(x)=x^3, \forall$ $x\in \mathbb R $.

(ii) $x+y=0$ is a tangent to the graph of $f$.

If we integrate (i), we get $f(x)=\frac{1}{4}x^4+C$. As for (ii), I assume this has something to do with a circle or $-x$ but I'm not understanding how to solve for this without having issues as far as differentiating wrt x. Maybe pick an $(x_o,y_o)$ coordinate or something?

(I'm probably missing something totally obvious, but its finals week so...)

Answer 1

$\int{f'(x)}dx=\int{x^3}dx=\tfrac{1}{4}x^4+c$

Then notice that $x+y=0$ implies $y=-x$ is a tangent line for your curve.

Now try solving this by graphing. https://www.desmos.com/calculator is a good resource I prefer.

Answer 2

(i) $\int{f'(x)}dx=\int{x^3}dx=\tfrac{1}{4}x^4+C$

(ii) $x+y=0\Rightarrow y=-x$

$$\frac{dy}{dx}=-1 \Rightarrow x^3=-1$$

$$x=-1 \Rightarrow y=1$$

$$y=\frac14x^4+C\Rightarrow 1=\frac14(-1)^4+C$$

$$C=\frac34$$

A solution for $f$ is $f(x)=\frac14x^4+\frac34$. Thanks for the help!