جواب فعّالیت صفحه 49 درس 2 ریاضی و آمار یازدهم انسانی (تابع)

\(\begin{array}{l}{f_3}(x) = {f_1}(x) + {f_2}(x) = ({x^2} - 1) + (x + 1) = {x^2} + x\\\\{f_4}(x) = {f_1}(x) - {f_2}(x) = ({x^2} - 1) - (x + 1) = {x^2} - x - 2\\\\{f_5}(x) = {f_2}(x) - {f_1}(x) = (x + 1) - ({x^2} - 1) = - {x^2} + x + 2\\\\{f_6}(x) = {f_1}(x) \times {f_2}(x) = ({x^2} - 1)(x + 1) = {x^3} + {x^2} - x - 1\\\\{f_7}(x) = \frac{{{f_1}(x)}}{{{f_2}(x)}} = \frac{{({x^2} - 1)}}{{(x + 1)}} = \frac{{(x - 1)(x + 1)}}{{(x + 1)}} = x - 1\\\\{f_8}(x) = \frac{{{f_2}(x)}}{{{f_1}(x)}} = \frac{{(x + 1)}}{{({x^2} - 1)}} = \frac{{(x + 1)}}{{(x - 1)(x + 1)}} = \frac{1}{{x - 1}}\end{array}\)

پس از محاسبه توابع در کفه های ترازو داریم :

سمت چپ ترازو :

\(\begin{array}{l}{f_6}(2) = {2^3} + {2^2} - 2 - 1 = 9\\\\{f_1}(2) + {f_2}(2) = 6\\\\|{f_4}(2)| = 0\\\\|{f_2}(2) - {f_1}(2)| = 0\\\\{f_7}(2) = 2 - 1 = 1\\\\\frac{{{f_2}(2)}}{{{f_1}(2)}} = \frac{{2 + 1}}{{4 - 1}} = \frac{3}{3} = 1\end{array}\)

کفه سمت چپ \(= 9 + 6 + 0 + 0 + 1 + 0 + 1 = 17\)

سمت راست ترازو :

\(\begin{array}{l}|{f_1}(2) - {f_2}(2)| = |{2^2} - 2 - 2| = 0\\\\|{f_5}(2)| = | - {2^2} + 2 + 2| = 0\\\\|{f_1}(2) - {f_2}(2)| = 0\\\\{f_3}(2) = {2^2} + 2 = 6\\\\{f_8}(2) = \frac{1}{{2 - 1}} = \frac{1}{1} = 1\end{array}\)

کفه سمت راست \(= 0 + 9 + 1 + 1 + 0 + 6 = 17\)

دو کفه با هم برابرند.

نتیجه می گیریم:

\(\begin{array}{l}{f_6}(2) = {f_1}(2) \times {f_2}(2)\\\\{f_3}(2) = {f_1}(2) + {f_2}(2)\\\\|{f_2}(2) - {f_1}(2)| = |{f_1}(2) - {f_2}(2)|\\\\|{f_4}(2)| = |{f_5}(2)|\\\\{f_7}(2) = \frac{{{f_1}(2)}}{{{f_2}(2)}}\\\\{f_8}(2) = \frac{{{f_2}(2)}}{{{f_1}(2)}}\end{array}\)