جواب کار در کلاس صفحه 74 درس 4 ریاضی دوازدهم تجربی (مشتق)

الف

\(\begin{array}{l}f(8) = - {(8)^2} + 10(8) = 16\\\\ \Rightarrow f'(8) = \mathop {\lim }\limits_{x \to 8} \frac{{f(x) - f(8)}}{{x - 8}} = \mathop {\lim }\limits_{x \to 8} \frac{{ - {x^2} + 10x - 16}}{{x - 8}}\\\\ = \mathop {\lim }\limits_{x \to 8} \frac{{ - (x - 2)(x - 8)}}{{x - 8}} = - 6\\\\f(5) = - {(5)^2} + 10(5) = 25\\\\ \Rightarrow f'(5) = \mathop {\lim }\limits_{x \to 5} \frac{{f(x) - f(5)}}{{x - 5}} = \mathop {\lim }\limits_{x \to 5} \frac{{ - {x^2} + 10x - 25}}{{x - 5}} = \mathop {\lim }\limits_{x \to 5} \frac{{ - {{(x - 5)}^2}}}{{x - 5}}\\\\ = \mathop {\lim }\limits_{x \to 5} - (x - 5) = 0\end{array}\)

ب

A و F

پ

در نقاط A  و E  و D، شیب مثبت و در نقاط B و G و F شیب منفی است.

ت

\(f'(3) > f'(4)\) و \({m_E} > {m_D}\)

ث

\(\begin{array}{l}f(3) = - {(3)^2} + 10(3) = 21\\\\ \Rightarrow f'(3) = \mathop {\lim }\limits_{x \to 3} \frac{{f(x) - f(3)}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \frac{{ - {x^2} + 10x - 21}}{{x - 3}}\\\\ = \mathop {\lim }\limits_{x \to 3} \frac{{ - (x - 7)(x - 3)}}{{x - 3}} = 4\\\\f(4) = - {(4)^2} + 10(4) = 24\\\\ \Rightarrow f'(4) = \mathop {\lim }\limits_{x \to 4} \frac{{f(x) - f(4)}}{{x - 4}} = \mathop {\lim }\limits_{x \to 4} \frac{{ - {x^2} + 10x - 24}}{{x - 4}}\\\\ = \mathop {\lim }\limits_{x \to 4} \frac{{ - (x - 6)(x - 4)}}{{x - 4}} = \mathop {\lim }\limits_{x \to 4} - (x - 6) = 2\end{array}\)

لذا \(f'(3) > f'(4)\)