Question

How to find a function f such that f′(x)=3x^3 and the line x+y=0 is tangent to the graph of f ?

Answer 1

It's messy, but I got `f(x) = 3/4x^4+(root(3)9)/4`

Explanation:

We need: `f'(x) = 3x^3`,

So we must have:

`f(x) = 3/4 x^4+C` for some constant `C`.

In order to be tangent to `y=-x`, we need an `a` with `f'(a) = -1`.

Solving `3x^3=-1`, we get `x=-1/root(3)3`.

So we want `y=-x` to be tangent at `a = -1/root(3)3`.

Now we find `f(a) = f(-1/root(3)3)`

`f(-1/root(3)3) = 3/4 1/(3root(3)3) + C = 1/(4root(3)3) = root(3)9/12 +C`

At the point `(-1/root(3)3, (root(3)9)/12+C)`, the tangent line is:

`y-((root(3)9)/12+C) = -1(x-(-1/root(3)3))`

`y-((root(3)9)/12+C)= -1x-1/root(3)3`

`y = -1x-1/root(3)3 + ((root(3)9)/12+C)`

`y = -1x - (4root(3)9)/12 + (root(3)9)/12+C`

`y = -x -(3root(3)9)/12 +C`

`y = -x - root(3)9/4 +C`

In order for the tangent line to be `y=-x`, we must have `C = root(3)9/4`

Now that we've finished, let's look at the graph.

(Image using Desmos https://www.desmos.com/calculator )

Here is Socratic's graphing utility graph.

graph{(y-3/4x^4-(root(3)9)/4)(x+y)=0 [-3.158, 1.708, -0.457, 1.976]}