Question 79·Hard·Advanced Math

$f(1)=7$ means the function gives $7$ when the input is $1$. A negative exponent means a reciprocal, so:

Type $t_1+t_1^{-1}\sim7\{t_1>1\}$. The $\sim$ tells Desmos to find a value that makes both sides equal; $t_1$ names the base it finds. Under PARAMETERS, Desmos shows $t_1\approx6.8541$. The restriction $t_1>1$ selects the base the question allows.

Type $f(x)=t_1^x+t_1^{-x}$ beneath the regression. Desmos uses the value it found for $t_1$, so this rule keeps the same base in both terms without rounding it yourself.

Type $f(3)$ on the next line. Desmos shows $322$. The base was an intermediate value; the requested function output is what you grid in. So $f(3)=322$. Grid in 322.

Let $a=t$ and $b=t^{-1}$. Their product is $1$, since $ab=t\cdot t^{-1}=1$. The given $f(1)=7$ says their sum is $a+b=7$.

Use the cube-of-a-sum identity:

The last term accounts for the mixed products in the cube. Dropping it would make $(a+b)^3$ too large for the sum of cubes.

Subtract the mixed term:

Here $a^3+b^3=f(3)$. Put in the sum $7$ and product $1$:

Type $7^3-3(1)(7)$ into Desmos. It shows $322$, so the value of $f(3)$ is $322$. Grid in 322.