Answer
The curves are tangent to each other at the point $(2,6)$
We can see a sketch of the graphs below.
Work Step by Step
$y = x^3-3x+4$
$\frac{dy}{dx} = 3x^2-3$
$y = 3(x^2-x) = 3x^2-3x$
$\frac{dy}{dx} = 6x-3$
If the two curves are tangent to each other at a point, then the slopes of the graphs must be equal at that point.
We can find the points where the slopes are equal.
$3x^2-3 = 6x-3$
$3x^2-6x = 0$
$3x(x-2) = 0$
$x=0$ or $x=2$
When $x=0$:
$x^3-3x+4 = (0)^3-3(0)+4 = 4$
$3(x^2-x)= 3(0^2-0) = 0$
Since the values are not equal at $x=0$, the curves are not tangent at $x=0$
When $x=2$:
$x^3-3x+4 = (2)^3-3(2)+4 = 6$
$3(x^2-x)= 3(2^2-2) = 6$
Since the values are equal at $x=2$, and the slopes are equal at $x=2$, the curves are tangent at $x=2$.
Therefore, the curves are tangent to each other at the point $(2,6)$.
We can see a sketch of the graphs below.