Prove that $g$ is tangent to the graph of $f$

Question

So I had these two functions and the following exercise:

$f(x)=x^3-2x^2$

$g_p(x)=px$

Prove that $g_{-1}$ is a tangent line to the graph of $f$. How many points do $g_{-1}$ and $f(x)$ have in common?

So yeah, the points in common are fairly simple to figure out:

Points in common: $x=1$ and $x=0$

So to see if it's actually tangent to the graph, I took the derivative of both functions at the points where they meet.

The result:

$f'(x)=3x^2-4x$

$g'_{-1}(x)=-1$

So $f'(1)=-1$ and $f'(0)=0$

So I'm not sure why I get this answer? By the question, we are supposed to prove that its tangent to the graph of $f$ but by my calculation, it seems like only $f(1)$ is tangent with $g_{-1}(1)$.

Help is appreciated! (Please avoid hint answers).

Answer 1

You have obtained exactly the right answer, you have just misinterpreted the requirements of the question.

To say that $g_{-1}$ is tangent to $f$ means that it is tangent in at least one place. It does not have to be tangent every time the two functions meet.

You have shown that the two functions are tangent at $x=1$, and also shown that $g_{-1}$ meets $f$ at one other place, so you have answered both parts of the question - well done!

Answer 2

Two function touches each other in a common point if the derivatives in this common point are the same. You have found that $g_{-1}$ is tangent to $f$ in $x=1$. Note that this tangent in $x=1$ may have other common points with $f$, in which it is not tangent. Being tangent is a local attribute.