명제와 조건 연습문제 — Sundstrom, Mathematical Reasoning · 1.1절 · 6c번

Sundstrom, Mathematical Reasoning · 1.1절 · 6c번영어 원문

Following is a statement of a theorem which can be proven using calculus or precalculus mathematics. For this theorem, $a$ , $b$ , and $c$ are real numbers.

Theorem If $f$ is a quadratic function of the form $f\left( x \right) = ax^2 + bx + c$ and $a \lt 0$ , then the function $f$ has a maximum value when $x = \dfrac{{ - b}}{{2a}}$ .

Using only this theorem, what can be concluded about the functions given by the following formulas?

$k\left( x \right) = 8x^2 - 5x - 7$

정답·풀이 보기 (영어 원문)

No conclusion can be made about this function from this theorem.

한 번 더 — 비슷한 문제Sundstrom, Mathematical Reasoning · 2.2절 · 7a번영어 원문

Use previously proven logical equivalencies to prove each of the following logical equivalencies about conditionals with conjunctions:

$\left[ {\left( {P \wedge Q} \right) \to R} \right] \equiv \left( {P \to R} \right) \vee \left( {Q \to R} \right)$

다 풀었으면 정답 확인

In this case, it may be better to work with the right side first.

$\begin{aligned}\left( {P \to R} \right) \vee \left( {Q \to R} \right) & \equiv \left( \neg P \vee R \right) \vee \left( \neg Q \vee R \right) \\ & \equiv \left( \neg P \vee \neg Q \right) \vee \left( R \vee R \right) \\ & \equiv \left( \neg P \vee \neg Q \right) \vee R \\ & \equiv \neg \left( P \wedge Q \right) \vee R \\ & \equiv \left( P \wedge Q \right) \to R\end{aligned}$

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