In $g(x)=-2\,f\bigl(\frac{x}{2}-3\bigr)+5$, what is the "input" being fed into $f$? How does the negative factor in front of $f$ affect whether a minimum of $f$ becomes a maximum or minimum of $g$?

Once you know the input where $f$ is minimized, set the expression $\dfrac{x}{2}-3$ equal to that value and solve for $x$.

From your equation involving $\dfrac{x}{2}-3$, isolate $x$ step by step: undo the subtraction, then undo the division.

In Desmos, type f(x)=3x^2-12x+7 to graph the function $f$.

On a new line, type g(x)=-2*f(x/2-3)+5 to graph $g(x)$ using your definition of $f$.

Click on the graph of $g(x)$; Desmos will show a point at its highest point (the vertex). Note the x-coordinate of this maximum point—that is the value of $x$ where $g(x)$ attains its maximum.

The function

is a quadratic with $a=3>0$, so it opens upward and has a minimum at its vertex.

The $x$-coordinate of the vertex of a quadratic $ax^2+bx+c$ is

Here, $a=3$ and $b=-12$, so

So $f(x)$ has its minimum value when its input is $x=2$.

The function $g$ is defined by

Notice two important things:

• The input to $f$ is $\dfrac{x}{2}-3$ instead of $x$.
• There is a factor of $-2$ in front, which flips the graph vertically.

Because of the negative factor, when $f$ gets smaller, $g$ gets larger (the $-2$ changes a small value of $f$ into a large value of $g$ after adding $5$).

So to make $g(x)$ as large as possible (its maximum), we want $f\bigl(\dfrac{x}{2}-3\bigr)$ to be as small as possible — that is, we want the input $\dfrac{x}{2}-3$ to be the $x$-coordinate of the minimum of $f$, which is $2$ from Step 1.

Let the input to $f$ in $g(x)$ be

We want $u$ to equal $2$, since $f$ is minimized at input $2$. So we set up the equation

Now solve this equation for $x$.

Solve

Add $3$ to both sides:

Multiply both sides by $2$:

When $x=10$, the input to $f$ is

which makes $f$ as small as possible and therefore makes $g(x)$ as large as possible.

Thus, $g(x)$ attains its maximum when $x=10$.