Let the domain of the function f(x) = cos⁻¹((4x+5)/(3x-7)) be [α, β] and the domain of g(x) = log₂(2

Q: Why can't I just cross-multiply the denominator in the inequality $\frac{4x+5}{3x-7} \le 1$?

This is a classic trap! Cross-multiplying is only safe if you know the sign of the denominator. Since $3x-7$ can be positive or negative depending on the value of $x$, multiplying by it could flip the inequality sign, leading to an incorrect answer. Always bring terms to one side to form a single fraction and use the Wavy Curve Method to analyze the sign changes safely.

Q: Why is the interval open at $7/3$ in the domain of $f(x)$?

Excellent observation. Look at the expression $\frac{4x+5}{3x-7}$. If $x = 7/3$, the denominator becomes zero, making the expression undefined. Even though the inequality is $\le$, we must exclude any point that causes division by zero. In mathematics, the domain is the set of all inputs where the function is well-defined, and division by zero is strictly forbidden.

Q: Why didn't the inequality sign flip when we removed the logarithm in $g(x)$?

That depends entirely on the base of the logarithm. The property states that for $\log_b(u) 1$, the inequality remains the same ($u < b^c$). If $0 < b < 1$, the inequality flips. Since our base is $27$, which is greater than $1$, the function is monotonically increasing, so the inequality sign is preserved. You are doing great—keep paying attention to those bases!

Question

Let the domain of the function $f(x) = \cos^{-1}(\frac{4x+5}{3x-7})$ be $[\alpha, \beta]$ and the domain of $g(x) = \log_2(2 - 6 \log_{27}(2x + 5))$ be $(\gamma, \delta)$. Then $|7(\alpha + \beta) + 4(\gamma + \delta)|$ is equal to ________

Sigma Percentile · Accepted Answer

### Analyzing the Setup The domain of a function defines the set of all possible input values for which the function is real-valued. We are tasked with finding the domains of two functions: $f(x) = \cos^{-1}\left(\frac{4x+5}{3x-7}\right)$ and $g(x) = \log_2(2 - 6 \log_{27}(2x + 5))$. We define the domain of $f(x)$ as $[\alpha, \beta]$ and the domain of $g(x)$ as $(\gamma, \delta)$. Our final objective is to calculate the value of $|7(\alpha + \beta) + 4(\gamma + \delta)|$. ### Phase 1: The Inverse Trigonometric Constraint The function $f(x) = \cos^{-1}(u)$ is defined only when the input $u$ satisfies $-1 \le u \le 1$. Therefore, we must solve the double inequality: $$-1 \le \frac{4x+5}{3x-7} \le 1$$ We split this into two separate inequalities to solve them systematically. **Part A: Solving $\frac{4x+5}{3x-7} \le 1$** Subtracting $1$ from both sides, we obtain: $$\frac{4x+5}{3x-7} - 1 \le 0 \implies \frac{4x+5 - (3x-7)}{3x-7} \le 0 \implies \frac{x+12}{3x-7} \le 0$$ Using the Wavy Curve Method with critical points $x = -12$ and $x = 7/3$, we find the solution set: $$x \in [-12, 7/3)$$ **Part B: Solving $\frac{4x+5}{3x-7} \ge -1$** Adding $1$ to both sides, we obtain: $$\frac{4x+5}{3x-7} + 1 \ge 0 \implies \frac{4x+5 + 3x-7}{3x-7} \ge 0 \implies \frac{7x-2}{3x-7} \ge 0$$ Using the Wavy Curve Method with critical points $x = 2/7$ and $x = 7/3$, we find the solution set: $$x \in (-\infty, 2/7] \cup (7/3, \infty)$$ **Intersection for $f(x)$** The intersection of $[-12, 7/3)$ and $(-\infty, 2/7] \cup (7/3, \infty)$ is $[-12, 2/7]$. Thus, we identify: $$\alpha = -12, \quad \beta = 2/7$$ ### Phase 2: The Logarithmic Puzzle For $g(x) = \log_2(2 - 6 \log_{27}(2x + 5))$, we must satisfy two conditions for the logarithm to be defined. First, the inner argument must be positive: $$2x + 5 > 0 \implies x > -5/2$$ Second, the argument of the outer logarithm must be positive: $$2 - 6 \log_{27}(2x + 5) > 0 \implies \log_{27}(2x + 5) < 1/3$$ Exponentiating both sides with base $27$: $$2x + 5 < 27^{1/3} \implies 2x + 5 < 3 \implies 2x < -2 \implies x < -1$$ Combining $x > -5/2$ and $x < -1$, the domain is $(-5/2, -1)$. Thus: $$\gamma = -5/2, \quad \delta = -1$$ ### Phase 3: Final Calculation We now substitute our identified values $\alpha = -12$, $\beta = 2/7$, $\gamma = -5/2$, and $\delta = -1$ into the required expression: $$|7(\alpha + \beta) + 4(\gamma + \delta)| = \left| 7\left(-12 + \frac{2}{7}\right) + 4\left(-\frac{5}{2} - 1\right) \right|$$ Expanding the terms: $$\left| 7(-12) + 7\left(\frac{2}{7}\right) + 4\left(-\frac{5}{2}\right) + 4(-1) \right| = | -84 + 2 - 10 - 4 |$$ Simplifying the sum inside the modulus: $$| -96 | = 96$$ The final result is **96**.

Animated Solution for Mathematics - Functions: Let the domain of the function $f (x) = cos^{- 1} (\frac{4 x + 5}{3 x - 7})$ be $[α, β]$ and the domain of $g (x) = lo g_{2} (2 - 6 lo g_{27} (2 x + 5))$ be $(γ, δ)$ . Then $∣7 (α + β) + 4 (γ + δ) ∣$ is equal to ________

Enter Numerical Value:

Visualized Solution

$P r o b l e m O v er v i e w$

$D o main C o n d i t i o n f or cos^{- 1} (u)$

$S o l v in g t h e R i g h t I n e q u a l i t y$

$W a v y C u r v e M e t h o d (P a r t 1)$

$S o l v in g t h e L e f t I n e q u a l i t y$

$W a v y C u r v e M e t h o d (P a r t 2)$

$D o main o f f (x)$

$D o main C o n d i t i o n s f or g (x)$

$S o l v in g C o n d i t i o n 1$

$S o l v in g C o n d i t i o n 2$

$D o main o f g (x)$

$S u b s t i t u t in g V a l u es$

$F ina l E v a l u a t i o n$

The Sigma Insight: Domain and Range of a Function

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Problem Overview

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Domain Condition for cos−1(u)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving the Right Inequality

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Wavy Curve Method (Part 1)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving the Left Inequality

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Wavy Curve Method (Part 2)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Domain of f(x)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Domain Conditions for g(x)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving Condition 1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving Condition 2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Domain of g(x)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Evaluation