Notice that both $g(1)$ and $g(2)$ can be related to the expression $r+\dfrac{1}{r}$. Try letting $t=r+\dfrac{1}{r}$ and rewrite $r^2+\dfrac{1}{r^2}$ in terms of $t$.

Remember that $(r+1/r)^2=r^2+2+1/r^2$, which lets you express $r^2+1/r^2$ in terms of $t$. Later, for $g(3)$, consider using the formula for $a^3+b^3$ in terms of $a+b$.

After solving the quadratic for $t$, think about which value makes sense given that $r$ is positive. Then use your $t$ value in the cube identity to find $r^3+1/r^3$.

In Desmos, type y = x^2 - x - 30. The $x$-intercepts of this parabola are the solutions for $t$. Identify the positive intercept, since $t=r+1/r$ must be positive when $r>0$.

In a new Desmos line, enter f(t) = t^3 - 3*t and then evaluate it at the positive value of $t$ you found (for example by typing f(<your t-value>)). The output is the value of $g(3)$.

We are given $g(x)=r^x+r^{-x}$.

Compute the first two values:

• $g(1)=r^1+r^{-1}=r+\dfrac{1}{r}$
• $g(2)=r^2+r^{-2}$

The equation $g(2)-g(1)=28$ becomes

Let $t=r+\dfrac{1}{r}$. Then $g(1)=t$.

Now express $r^2+\dfrac{1}{r^2}$ in terms of $t$:

so

Thus $g(2)=r^2+r^{-2}=t^2-2$, and the equation $g(2)-g(1)=28$ becomes

Simplify the equation:

Factor the quadratic:

So $t=6$ or $t=-5$.

Since $r>0$, both $r$ and $1/r$ are positive, so $t=r+1/r$ must be positive. Therefore $t=-5$ is impossible, and we must have

We want $g(3)=r^3+r^{-3}$.

Use the identity for the sum of cubes: for $a$ and $b$,

Here $a=r$ and $b=\dfrac{1}{r}$, so $ab=1$ and $a+b=t$.

Thus

Substitute $t=6$ into $t^3-3t$:

So the value of $g(3)$ is 198.