Find all tangents to $f(x)$, so area between that tangent, $f(x)$ and $x$-axis will be $\frac{3}{4}$

Question

Find all tangents to a function $f(x) = x^3$, so area between found tangent, $f(x)$ and x-axis is equal to $\frac{3}{4}$.

I assume that area between $f(x)$, x-axis and tangent $ax + b$ will be same as shifted tangent $ax - b$. And point of intersection with x-axis at $\frac{-b}{a}$.

So I tried to put it into system of equations:

$\int_0^\frac{-b}{a} x^3 - (ax + b) dx = \frac{3}{4}$

$\int_\frac{-b}{a}^0 x^3 - (ax - b) dx = \frac{3}{4}$

Which doesn't work quite well because I'm unable to involve the fact that $ax \pm b$ is tangent.

Any tips how to continue or where to start right way?

Thanks a lot for any help.

Answer 1

Let $x_0\geq 0$ then the tangent at $x_0$ to the graph of $x^3$ is $y=3x_0^2(x-x_0)+x_0^3$. Such tangent line intersects the $x$-axis at $2x_0/3$. Hence the desired area is $$\int_{0}^{2x_0/3}x^3dx+\int_{2x_0/3}^{x_0}(x^3-(3x_0^2(x-x_0)+x_0^3))\,dx.$$ which is equal to $$\int_{0}^{x_0}x^3dx-\int_{-x_0/3}^{0}(3x_0^2t+x_0^3)\,dt=\frac{x_0^4}{12}.$$ Can you take it from here?