The problem has three parts: (a) Express the quadratic $y = 5 - 2x - 4x^2$ in the form $A - B(x+C)^2$, where $A$, $B$, and $C$ are constants. Then, find the maximum value of the function $f(x) = 5 - 2x - 4x^2$. (b) Given $3 \log_2 r = n$ and $\log_2 (4m) = n+4$, find the values of $m$ and $n$. Note that in the first equation $r=1$. (c) The functions $f$ and $g$ are defined by $f(x) = x^2 + 1$ and $g(x) = x - 2$. Find the value of $x$ such that $f(g(x)) = 5$.

1. Problem Description

The problem has three parts:

(a) Express the quadratic $y = 5 - 2x - 4x^2$ in the form $A - B(x+C)^2$, where $A$, $B$, and $C$ are constants. Then, find the maximum value of the function $f(x) = 5 - 2x - 4x^2$.

(b) Given $3 \log_2 r = n$ and $\log_2 (4m) = n+4$, find the values of $m$ and $n$. Note that in the first equation $r=1$.

(c) The functions $f$ and $g$ are defined by $f(x) = x^2 + 1$ and $g(x) = x - 2$. Find the value of $x$ such that $f(g(x)) = 5$.

2. Solution Steps

(a) Completing the square:

$y = 5 - 2x - 4x^2 = 5 - 4(x^2 + \frac{1}{2}x)$

$y = 5 - 4(x^2 + \frac{1}{2}x + (\frac{1}{4})^2 - (\frac{1}{4})^2)$

$y = 5 - 4((x + \frac{1}{4})^2 - \frac{1}{16})$

$y = 5 - 4(x + \frac{1}{4})^2 + \frac{1}{4}$

$y = \frac{21}{4} - 4(x + \frac{1}{4})^2$

Thus, $A = \frac{21}{4}$, $B = 4$, and $C = \frac{1}{4}$.

The maximum value occurs when $(x + \frac{1}{4})^2 = 0$, i.e., $x = -\frac{1}{4}$. The maximum value is $y = \frac{21}{4}$.

(b) We have $3 \log_2 1 = n$ and $\log_2 (4m) = n + 4$.

Since $\log_2 1 = 0$, we have $3(0) = n$, so $n = 0$.

Then, $\log_2 (4m) = 0 + 4 = 4$.

$4m = 2^4 = 16$

$m = \frac{16}{4} = 4$

(c) $f(x) = x^2 + 1$ and $g(x) = x - 2$.

$f(g(x)) = f(x - 2) = (x - 2)^2 + 1$

We are given $f(g(x)) = 5$, so $(x - 2)^2 + 1 = 5$.

$(x - 2)^2 = 4$

$x - 2 = \pm 2$

$x = 2 \pm 2$

So $x = 2 + 2 = 4$ or $x = 2 - 2 = 0$.

3. Final Answer

(a) $A = \frac{21}{4}$, $B = 4$, $C = \frac{1}{4}$. The maximum value is $\frac{21}{4}$.

(b) $m = 4$, $n = 0$.

(c) $x = 0, 4$.