Answer
$f(x)=\frac{x^{4}}{4}+\frac{3}{4}$
Work Step by Step
It is given that:
$$f'(x)=x^{3}$$
Integrate both sides with respect to $x$:
$$\int f'(x) dx=\int x^{3}dx$$
$$f(x)=\frac{x^{4}}{4}+c$$
Find $c$.
The slope of the tangent line to the graph of $f$ at the point $(x,f(x))$ is:
$$m=f'(x)$$
The slope intercept form of the line $x+y=0$ is $y=-x$.
The slope of the line $y=-x$ is $-1$.
So:
$$m=f'(x) \to -1=x^{3} \to x=-1$$
The $y$-coordinate of $x=-1$ is $y=-(-1)=1$.
So the tangent line touches the graph of $f$ at the point $(-1,1)$ which mean that $f(-1)=1$.
$$f(x)=\frac{x^{4}}{4}+c \to f(-1)=\frac{(-1)^{4}}{4}+c$$
$$1=\frac{(-1)^{4}}{4}+c$$
$$1=\frac{1}{4}+c$$
$$\frac{3}{4}=c$$
So the equation of $f$ is:
$$f(x)=\frac{x^{4}}{4}+c \to f(x)=\frac{x^{4}}{4}+\frac{3}{4} $$
