Question

How do you find the point where the graphs of `f(x)=x^3-2x` and `g(x)=0.5x^2-1.5` are tangent to each other?

Answer 1

I found `(1,-1)` although I have the strong sensation that there is an easier way!

Explanation:

Let us try deriving both of them! We should find the slope of the tangent...to both curves!!!!
`f'(x)=3x^2-2`
`g'(x)=2*1/2x=x`

Let us set them equal:
`3x^2-2=x`
Using the Quadratic Formula you get:
`x_(1,2)=(1+-sqrt(1+24))/6=`
`x_1=1`
`x_2=-2/3`
Apparently we have 2 possible `x` values where the two curves share the tangent with the same inclination. Let us use these values to find `y`:
`x_1=1` `y_1=-1` FOR BOTH
`x_2=-2/3` `y_(2a)=28/27` first curve
`y_(2b)=-23/18` for the second curve.
This means that at `x_2` the curves have the tangent with the same inclination but they do not touch each others!

Graphically you can see the tangents (in black):