DC CAPE Algebra I Building Functions. Practice it free.

The common difference is $d = -5$. With $a_1 = 40$, the $12$th term is $a_{12} = 40 + 11(-5) = -15$.

Set $y=(x-4)^{3}$. Swap: $x=(y-4)^{3}$. Take the cube root: $\sqrt[3]{x}=y-4$. Add $4$: $y=\sqrt[3]{x}+4$. So $f^{-1}(x)=\sqrt[3]{x}+4\textsf{.}$

Work from the outside in. Since $m(4) = 1$, we need $n(k(x)) = 4$. Since $n(4) = 4$, we need $k(x) = 4$. From the table, $k(2) = 4$, so $x = 2$.

Multiplying the outputs by $2$ doubles every $y$-value while leaving $x$-values unchanged, so the graph is stretched vertically by a factor of $2$.

Subtracting $5$ outside $f$ decreases every output by $5\textsf{.}$ The graph shifts down $5$ units.

Following the pattern:
1st term, or $n = 1\textsf{:}$ $7 \cdot 2^{0}$
2nd term, or $n = 2\textsf{:}$ $7 \cdot 2^{1}$
3rd term, or $n = 3\textsf{:}$ $7 \cdot 2^{2}$
4th term, or $n = 4\textsf{:}$ $7 \cdot 2^{3}$
Since each term has an exponent that is $1$ less than the term number, the $n$th term is $w = 7 \cdot 2^{n-1}$.

A left shift of $1$ replaces $x$ with $x + 1$. At the $x$-intercept, $y = 0$, so $7(x + 1) = 14$. Then $7x + 7 = 14$, $7x = 7$, and $x = 1$.

$2^{4} = 16$, so $\log_{2}(16) = 4$.