Key Concepts and Formulas

• Range of a Composite Function: To find the range of $f(g(x))$ over a domain for $x$, first find the range of $g(x)$ for that domain. This range of $g(x)$ becomes the domain for $f(x)$. Then, analyze the monotonicity of $f(x)$ over this new domain to determine the range of $f(g(x))$.
• Monotonicity using Derivatives: A function $h(x)$ is strictly increasing on an interval if $h'(x) > 0$ for all $x$ in the interval, and strictly decreasing if $h'(x) < 0$.
• Derivative of a Rational Function: For $h(x) = \frac{ax+b}{cx+d}$, $h'(x) = \frac{ad-bc}{(cx+d)^2}$.

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Step-by-Step Solution

Step 1: Determine the Range of the Inner Function $g(x)$

We are given the inner function $g(x) = \frac{2-3x}{1-x}$ and the domain for $x$ is $[2,4]$. First, we evaluate $g(x)$ at the endpoints of the domain:

• At $x=2$: $g(2) = \frac{2-3(2)}{1-2} = \frac{2-6}{-1} = \frac{-4}{-1} = 4$
• At $x=4$: $g(4) = \frac{2-3(4)}{1-4} = \frac{2-12}{-3} = \frac{-10}{-3} = \frac{10}{3}$

Next, we determine the monotonicity of $g(x)$ by calculating its derivative $g'(x)$. We can rewrite $g(x)$ as $g(x) = \frac{-3x+2}{-x+1}$. Using the formula for the derivative of a rational function $\frac{ax+b}{cx+d}$ which is $\frac{ad-bc}{(cx+d)^2}$, with $a=-3, b=2, c=-1, d=1$: $g'(x) = \frac{(-3)(1) - (2)(-1)}{(-x+1)^2} = \frac{-3+2}{(1-x)^2} = \frac{-1}{(1-x)^2}$ For $x \in [2,4]$, $(1-x)^2$ is always positive and non-zero. Since the numerator is $-1$, $g'(x) < 0$ for all $x \in [2,4]$. This indicates that $g(x)$ is strictly decreasing on the interval $[2,4]$. Since $g(x)$ is decreasing, its maximum value on $[2,4]$ is $g(2)=4$ and its minimum value is $g(4)=\frac{10}{3}$. Thus, the range of $g(x)$ for $x \in [2,4]$ is $\left[\frac{10}{3}, 4\right]$. This interval will serve as the domain for the outer function $f$.

Step 2: Determine the Monotonicity of the Outer Function $f(x)$

The outer function is $f(x) = \frac{2x+3}{5x+2}$. The domain for $f$ is the range of $g(x)$, which is $\left[\frac{10}{3}, 4\right]$. We find the derivative $f'(x)$ to determine its monotonicity. Using the derivative formula for a rational function with $a=2, b=3, c=5, d=2$: $f'(x) = \frac{(2)(2) - (3)(5)}{(5x+2)^2} = \frac{4-15}{(5x+2)^2} = \frac{-11}{(5x+2)^2}$ For $x \in \left[\frac{10}{3}, 4\right]$, $x$ is positive, so $5x+2$ is positive, and $(5x+2)^2$ is always positive and non-zero. Since the numerator is $-11$, $f'(x) < 0$ for all $x$ in this interval. This means $f(x)$ is strictly decreasing on the interval $\left[\frac{10}{3}, 4\right]$.

Step 3: Determine the Range of the Composite Function $f(g(x))$

The range of $f(g(x))$ is determined by applying the outer function $f$ to the range of the inner function $g(x)$, which is $\left[\frac{10}{3}, 4\right]$. Since $f(x)$ is a decreasing function on $\left[\frac{10}{3}, 4\right]$, the maximum value of $f(g(x))$ will occur at the minimum value of $g(x)$, and the minimum value of $f(g(x))$ will occur at the maximum value of $g(x)$. The range of $f(g(x))$ is given as $[\alpha, \beta]$.

• The minimum value $\alpha$ is $f(\text{maximum value of } g(x)) = f(4)$.
• The maximum value $\beta$ is $f(\text{minimum value of } g(x)) = f\left(\frac{10}{3}\right)$.

Let's calculate these values:

• For $\alpha = f(4)$: $\alpha = \frac{2(4)+3}{5(4)+2} = \frac{8+3}{20+2} = \frac{11}{22} = \frac{1}{2}$
• For $\beta = f\left(\frac{10}{3}\right)$: $\beta = \frac{2\left(\frac{10}{3}\right)+3}{5\left(\frac{10}{3}\right)+2} = \frac{\frac{20}{3}+3}{\frac{50}{3}+2} = \frac{\frac{20+9}{3}}{\frac{50+6}{3}} = \frac{\frac{29}{3}}{\frac{56}{3}} = \frac{29}{56}$ So, the range of $f(g(x))$ is $[\alpha, \beta] = \left[\frac{1}{2}, \frac{29}{56}\right]$.

Step 4: Calculate $\frac{1}{\beta-\alpha}$

We need to find the value of $\frac{1}{\beta-\alpha}$. First, calculate $\beta - \alpha$: $\beta - \alpha = \frac{29}{56} - \frac{1}{2}$ To subtract the fractions, we use a common denominator, which is 56: $\beta - \alpha = \frac{29}{56} - \frac{1 \times 28}{2 \times 28} = \frac{29}{56} - \frac{28}{56} = \frac{1}{56}$ Now, we compute $\frac{1}{\beta-\alpha}$: $\frac{1}{\beta-\alpha} = \frac{1}{\frac{1}{56}} = 56$

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Common Mistakes & Tips

• Direction of Monotonicity: Carefully determine if the outer function $f$ is increasing or decreasing on the range of the inner function $g$. If $f$ is decreasing, the endpoints of the range of $f(g(x))$ are swapped compared to if $f$ were increasing.
• Domain of $f$: Ensure that the domain of $f$ (which is the range of $g$) does not contain any vertical asymptotes of $f$. In this problem, the domain of $f$ is $\left[\frac{10}{3}, 4\right]$, and the vertical asymptote of $f(x)=\frac{2x+3}{5x+2}$ is at $x=-\frac{2}{5}$, which is outside this interval.
• Calculation Errors: Double-check fraction arithmetic, especially when dealing with complex fractions or finding common denominators.

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Summary

To find the range of the composite function $f(g(x))$, we first determined the range of the inner function $g(x)$ over its given domain $[2,4]$. We found that $g(x)$ is a decreasing function and its range is $\left[\frac{10}{3}, 4\right]$. This range then served as the domain for the outer function $f(x)$. We analyzed the monotonicity of $f(x)$ on $\left[\frac{10}{3}, 4\right]$ and found it to be decreasing. Due to the decreasing nature of $f$, the minimum and maximum values of $f(g(x))$ were obtained by applying $f$ to the maximum and minimum values of $g(x)$ respectively. This yielded the range $[\alpha, \beta] = \left[\frac{1}{2}, \frac{29}{56}\right]$. Finally, we calculated $\frac{1}{\beta-\alpha}$ to be 56.

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The final answer is $\boxed{56}$.