Let the domain of the function f(x)=cos−1(3x−74x+5) be [α,β] and the domain of g(x)=log2(2−6log7(2x+5)) be [γ,δ]. Then ∣7(α+β)+4(γ+δ)∣ is equal to _____
Get the complete, step-by-step math solution for: "Let the domain of the function f(x) = cos^{-1} (4x + 5)/(3x - 7) be [α, β] and the domain of g(x) = _2(2 - 6 _7(2x + 5)) be [ ]. Then |7(α + β) + 4( +...". Powered by SolveForX AI math tutor.
Step-by-Step Solution
Step 1: Determine the domain of f(x)
For the inverse cosine function cos−1(u) to be defined, its argument u must satisfy −1≤u≤1. Therefore, we set up the inequality −1≤3x−74x+5≤1.
Step 2: Solve the inequality for f(x)
We solve the two inequalities separately. First, 3x−74x+5≥−1 simplifies to 3x−77x−2≥0. The critical points are x=72 and x=37. Analyzing the sign changes, we get x∈(−∞,72]∪(37,∞). Second, 3x−74x+5≤1 simplifies to 3x−7x+12≤0. The critical points are x=−12 and x=37. Analyzing the sign changes, we get x∈[−12,37). The intersection of these two solution sets gives the domain of f(x) as [−12,72]. Thus, α=−12 and β=72.
Step 3: Determine the domain of g(x)
For the logarithm function logb(u) to be defined, its argument u must be positive. First, for log2(2−6log7(2x+5)), we need 2−6log7(2x+5)>0. This simplifies to log7(2x+5)<31. Second, for log7(2x+5), we need 2x+5>0, which means x>−25. Combining log7(2x+5)<31 with the base 7 (which is greater than 1), we get 2x+5<71/3, so x<271/3−5. The intersection of x>−25 and x<271/3−5 gives the domain of g(x) as (−25,271/3−5). Thus, γ=−25 and δ=271/3−5.
Step 4: Calculate the final expression
Now we substitute the values of α,β,γ,δ into the expression ∣7(α+β)+4(γ+δ)∣. First, calculate α+β=−12+72=−782. Next, calculate γ+δ=−25+271/3−5=271/3−10. Substitute these sums into the given expression: 7(−782)+4(271/3−10). Simplify the expression to get ∣−82+2(71/3−10)∣=∣−82+2⋅71/3−20∣=∣−102+2⋅71/3∣.