If $g(x)= -\frac 12 ( \frac 19 x^2 -4)^2$ is obtained by doing $ g(x)=f( kx) $ ( horizontal stretch of function $f$), what is number $k$?

Question

Let function $f$ be defined by : $f(x)= -\frac 12 ( x^2 -4)^2$

Suppose that function $g(x)$ is $f(kx)$ ( horizontal strech of function $f$)

and that

$g(x)= -\frac 12 ( \frac 19 x^2 -4)^2$.

What is number $k$?

I would have said that $k=\frac 19$.

But apparently $k= \frac 13$.

How to explain this?

The image below shows that functon $g$ (in blue) and $h$ ( in green) are identical.

Answer 1

We are given $f(x)=-\dfrac12(x^2-4)^2$. Therefore, $f(kx)=-\dfrac12((kx)^2-4)^2=-\dfrac12(k^2x^2-4)^2$.

Since $g(x)=-\dfrac12(\dfrac19x^2-4)^2$, it follows that $k^2=\dfrac19$ and $k=\pm\dfrac13$.

Answer 2

$f(\frac{x}{3})=-\frac{1}{2}((\frac{x}{3})^2-4)^2=-\frac{1}{2}(\frac{x^2}{9}-4)^2=g(x)$