Сложно заданные функции |
Логические схемы |
$$\quad$$ $$y = f(x) = \log _{\frac{1}{4}} \left( {60 + 4x - x^2 } \right)$$ |
$$\quad$$ $$E(f) = \log _{\frac{1}{4}} \left( {E(60 + 4x - x^2 )} \right) $$ $$= \log _{\frac{1}{4}} \left( {E = ( - \infty ;64)} \right) $$ $$= ( - \infty ;\log _{\frac{1}{4}} 64) $$ $$= ( - \infty ; - 3)$$ |
$$\quad$$ $$y = f(x) = \frac{6}{{x^2 + 4x + 6}}$$ |
$$\quad$$ $$E(f) = \frac{6}{{E(x^2 + 4x + 6)}} $$ $$= \frac{6}{{E = \left[ {2; + \infty } \right)}} $$ $$= \left( {0;3} \right]$$ |
$$\quad$$ $$y = f(x) = 2^{x^2 + 2x} $$ |
$$\quad$$ $$E(f) = 2^{E(x^2 + 2x)} $$ $$= 2^{E = \left[ { - 1; + \infty } \right)} $$ $$= \left[ {2^{ - 1} ; + \infty } \right) $$ $$= \left[ {\frac{1}{2}; + \infty } \right)$$ |
$$\quad$$ $$y = f(x) = \frac{1}{{1 + \sqrt {10x - x^2 - 16} }}$$ |
$$\quad$$ $$E(f) = \frac{1}{{1 + \sqrt {E(10x - x^2 - 16)} }} $$ $$= \frac{1}{{1 + \sqrt {E = \left( { - \infty ;9} \right]} }} $$ $$= \frac{1}{{1 + \left[ { 0 ;\sqrt 9 } \right]}} $$ $$= \frac{1}{{1 + \left[ { 0 ;3} \right]}} $$ $$= \frac{1}{{\left[ { 1 ;4} \right]}} $$ $$= \left[ { \frac{1}{4};1} \right]$$ |
$$\quad$$ $$y = f(x) = 3\sqrt {\cos x} $$ |
$$\quad$$ $$E(f) = 3\sqrt {E(\cos x)} $$ $$= 3\sqrt {E = \left[ { - 1;1} \right]} $$ $$= 3\sqrt {\left[ {0;1} \right]} $$ $$= \left[ {0;3} \right]$$ |
$$\quad$$ $$y = f(x) = \cos \left( {tgx} \right),\;x \in \left[ {0;\frac{\pi }{4}} \right]$$ |
$$\quad$$ $$E(f) = \cos \left( {E\left( {tgx,x \in \left[ {0;\frac{\pi }{4}} \right]} \right)} \right) $$ $$= \cos \left( {E = \left[ {tg0;tg\frac{\pi }{4}} \right]} \right) $$ $$= \cos \left( {E = \left[ {0;1} \right]} \right) $$ $$= \left[ {\cos 0;\cos 1} \right] $$ $$= \left[ {0;\cos 1} \right]$$ |