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Home » 第二步：计算 $ f(f(3)) = f(-1.5) $ - DNSFLEX

第二步：计算 $ f(f(3)) = f(-1.5) $ - DNSFLEX

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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📌 Solution: The maximum height occurs at the vertex. For $ y = -x^2 + 6x - 8 $, $ a = -1 $, $ b = 6 $, so $ x = -\frac{6}{2(-1)} = 3 $. Substitute $ x = 3 $: $ y = -(3)^2 + 6(3) - 8 = -9 + 18 - 8 = 1 $. The maximum height is $ 1 $ unit. \boxed{1}

📌 Question: Given $ f(2x + 1) = 4x^2 + 12x + 9 $, find $ f(x^2 - 3) $.

📌 Solution: Let $ u = 2x + 1 $, so $ x = \frac{u - 1}{2} $. Substitute into $ f(u) $: $ f(u) = 4\left(\frac{u - 1}{2}\right)^2 + 12\left(\frac{u - 1}{2}\right) + 9 = (u - 1)^2 + 6(u - 1) + 9 = u^2 - 2u + 1 + 6u - 6 + 9 = u^2 + 4u + 4 $. Thus, $ f(x^2 - 3) = (x^2 - 3)^2 + 4(x^2 - 3) + 4 = x^4 - 6x^2 + 9 + 4x^2 - 12 + 4 = x^4 - 2x^2 + 1 $. \boxed{x^4 - 2x^2 + 1}

📌 Question: Solve the inequality $ \frac{2x - 5}{x + 3} \geq 1 $ to determine the range of temperatures where a species thrives.

📌 Solution: Subtract 1: $ \frac{2x - 5}{x + 3} - 1 \geq 0 \Rightarrow \frac{2x - 5 - (x + 3)}{x + 3} \geq 0 \Rightarrow \frac{x - 8}{x + 3} \geq 0 $. Critical points at $ x = 8 $ and $ x = -3 $. Test intervals:

📌 For $ x < -3 $, test $ x = -4 $: $ \frac{-12}{-1} = 12 > 0 $.

📌 For $ -3 < x < 8 $, test $ x = 0 $: $ \frac{-8}{3} < 0 $.

📌 For $ x > 8 $, test $ x = 9 $: $ \frac{1}{12} > 0 $.

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The distance traveled by the train is given by the formula:
Distance = Speed × Time
Distance = 90 miles/hour × 2.5 hours = 225 miles
Perimeter = 2(Length + Width)
= 2(3w + w)
= 8w
Length = \( 3w = 24 \) meters
Area = Length × Width = 24 meters × 8 meters = 192 square meters

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