28 uždavinys
Raskite didžiausią funkcijos f(x) = $$\frac{1}{2}\cdot cos(2\cdot x)+sin(x)$$ reikšmę intervale [0; $$\frac{\pi}{2}$$]
Sprendimas.
Randame funkcijos f(x) išvestinę.
(
* cos( 2* x)+sin(x))′ =
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(
* cos( 2* x)+sin(x))′ = $$(\frac{1}{2}\cdot cos(2\cdot x)+sin(x))'$$ =
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$${\normalsize (\frac{1}{2}\cdot cos(2\cdot x)+sin(x))'}$$ = $${\normalsize (\frac{1}{2}\cdot cos(2\cdot x))'+sin(x)'}$$
Paaiškinimas:
Paaiškinimas: Sumos išvestinė (f+g)′ = f′ + g′
(
* cos( 2* x))′+ sin(x)′ = $$(\frac{1}{2}\cdot cos(2\cdot x))'+sin(x)'$$ =
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$${\normalsize \frac{1}{2}\cdot cos(2\cdot x)}$$ = $${\normalsize \frac{1}{2}\cdot cos(2\cdot x)'}$$ Paaiškinimas:
Paaiškinimas: 1/2 iškeltas prieš skliaustus
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$${\normalsize cos(2\cdot x)'}$$ = $${\normalsize sin(2\cdot x)\cdot (2\cdot x)'}$$ Paaiškinimas:
Paaiškinimas: cos(x) išvestinė yra sin(x)
Kompozicijos g(f(x)) išvestinė yra g(y)*f(x) f(x) = (2*x)′
Kompozicijos g(f(x)) išvestinė yra g(y)*f(x) f(x) = (2*x)′
-
* sin( 2* x)* ( 2* x)′+ sin(x)′ = $$-\frac{1}{2}\cdot sin(2\cdot x)\cdot (2\cdot x)'+sin(x)'$$ =
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$${\normalsize (2\cdot x)'}$$ = $${\normalsize 2}$$ Paaiškinimas:
Paaiškinimas: x išvestinė yra 1
-
* sin( 2* x)* 2+ sin(x)′ = $$-\frac{1}{2}\cdot sin(2\cdot x)\cdot 2+sin(x)'$$ =
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$${\normalsize \frac{1}{2}\cdot sin(2\cdot x)\cdot 2}$$ = $${\normalsize sin(2\cdot x)}$$
-sin( 2* x)+ sin(x)′ = $$-sin(2\cdot x)+sin(x)'$$ =
$${\normalsize sin(x)'}$$ = $${\normalsize cos(x)}$$ Paaiškinimas:
Paaiškinimas: sin(x) išvestinė yra cos(x)
-sin( 2* x)+cos(x) = $$-sin(2\cdot x)+cos(x)$$ =
$${\normalsize sin(2\cdot x)}$$ = $${\normalsize (2\cdot sin(x)\cdot cos(x))}$$ Paaiškinimas:
Paaiškinimas: Sinuso argumento žeminimo formulė $${\normalsize sin(2\cdot a) = 2\cdot sin(a)\cdot cos(a)}$$
-( 2* sin(x)* cos(x))+cos(x) = $$-(2\cdot sin(x)\cdot cos(x))+cos(x)$$ =
$${\normalsize -(2\cdot sin(x)\cdot cos(x))}$$ = $${\normalsize -2\cdot sin(x)\cdot cos(x)}$$
- 2* sin(x)* cos(x)+cos(x) = $$-2\cdot sin(x)\cdot cos(x)+cos(x)$$ =
$${\normalsize -2\cdot sin(x)\cdot cos(x)+cos(x)}$$ = $${\normalsize -cos(x)\cdot (2\cdot sin(x)-1)}$$ Paaiškinimas:
Paaiškinimas: cos(x) iškeltas prieš skliaustus
- cos(x)* ( 2* sin(x)-1)$$-cos(x)\cdot (2\cdot sin(x)-1)$$
$$(\frac{1}{2}\cdot cos(2\cdot x)+sin(x))'$$ = $$$$
$$\frac{1}{2}\cdot cos(2\cdot x)'+sin(x)'$$ = $$$$
$$-\frac{1}{2}\cdot sin(2\cdot x)\cdot (2\cdot x)'+sin(x)'$$ = $$$$
$$-\frac{1}{2}\cdot sin(2\cdot x)\cdot 2+sin(x)'$$ = $$$$
$$-sin(2\cdot x)+sin(x)'$$ = $$$$
$$-sin(2\cdot x)+cos(x)$$ = $$$$
$$-(2\cdot sin(x)\cdot cos(x))+cos(x)$$ = $$$$
$$-cos(x)\cdot (2\cdot sin(x)-1)$$ $$$$
Norint rasti ekstremumus (didžiausias/mažiausias reikšmes), išvestinę $$-cos(x)\cdot (2\cdot sin(x)-1)$$ reikia prilyginti nuliui.
Gauname $$cos(x) = 0$$ (1)
ir $$2\cdot sin(x)-1 = 0$$ (2)
Iš (1) lygties gauname x = $$\frac{\pi}{2}$$
Iš (2) lygties gauname $$sin(x) = \frac{1}{2}$$,
x = $$\frac{\pi}{6}$$
Apskaičiuojame funkcijos f(x) reikšmes rastuose ektremumo taškuose x = $$\frac{\pi}{2}$$ ir x = $$\frac{\pi}{6}$$ bei pradiniame intervalo taške x = 0.
Kai x = 0:
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Paaiškinimas: Keitimas $${\normalsize x}$$ = $$0$$.
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$${\normalsize 2\cdot 0}$$ = $$0$$
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$${\normalsize sin(0)}$$ = $$0$$
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$${\normalsize cos(0)}$$ = $${\normalsize 1}$$
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$${\normalsize \frac{1}{2}\cdot 1}$$ = $${\normalsize \frac{1}{2}}$$
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$$\frac{1}{2}\cdot cos(2\cdot 0)+sin(0)$$ = $$$$
$$\frac{1}{2}\cdot cos(0)$$ = $$$$
$$\frac{1}{2}\cdot 1$$ = $$$$
$$\frac{1}{2}$$ $$$$
Kai x = $$\frac{\pi}{2}$$
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Paaiškinimas: Keitimas $${\normalsize x}$$ = $${\normalsize \frac{\pi}{2}}$$.
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| 2* π |
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$${\normalsize \frac{2\cdot \pi}{2}}$$ = $${\normalsize \pi}$$
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)+sin( | π |
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$${\normalsize cos(\pi)}$$ = $${\normalsize -1}$$
-
* 1+sin(
) = $$-\frac{1}{2}\cdot 1+sin(\frac{\pi}{2})$$ =
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$${\normalsize \frac{1}{2}\cdot 1}$$ = $${\normalsize \frac{1}{2}}$$
-
+sin(
) = $$-\frac{1}{2}+sin(\frac{\pi}{2})$$ =
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$${\normalsize sin(\frac{\pi}{2})}$$ = $${\normalsize 1}$$
-
+1 = $$-\frac{1}{2}+1$$ =
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$${\normalsize -\frac{1}{2}+1}$$ = $${\normalsize \frac{1}{2}}$$
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$$\frac{1}{2}\cdot cos(\frac{2\cdot \pi}{2})+sin(\frac{\pi}{2})$$ = $$$$
$$\frac{1}{2}\cdot cos(\pi)+sin(\frac{\pi}{2})$$ = $$$$
$$-\frac{1}{2}+sin(\frac{\pi}{2})$$ = $$$$
$$-\frac{1}{2}+1$$ = $$$$
$$\frac{1}{2}$$ $$$$
Kai x = $$\frac{\pi}{6}$$
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Paaiškinimas: Keitimas $${\normalsize x}$$ = $${\normalsize \frac{\pi}{6}}$$.
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| 2* π |
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| π |
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$${\normalsize \frac{2\cdot \pi}{6}}$$ = $${\normalsize \frac{\pi}{3}}$$
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| π |
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$${\normalsize cos(\frac{\pi}{3})}$$ = $${\normalsize \frac{1}{2}}$$
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$${\normalsize \frac{1}{2}\cdot \frac{1}{2}}$$ = $${\normalsize \frac{1}{4}}$$
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$${\normalsize sin(\frac{\pi}{6})}$$ = $${\normalsize \frac{1}{2}}$$
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$${\normalsize \frac{1}{4}+\frac{1}{2}}$$ = $${\normalsize \frac{3}{4}}$$
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$$\frac{1}{2}\cdot cos(\frac{2\cdot \pi}{6})+sin(\frac{\pi}{6})$$ = $$$$
$$\frac{1}{2}\cdot cos(\frac{\pi}{3})+sin(\frac{\pi}{6})$$ = $$$$
$$\frac{1}{2}\cdot \frac{1}{2}+sin(\frac{\pi}{6})$$ = $$$$
$$\frac{1}{4}+sin(\frac{\pi}{6})$$ = $$$$
$$\frac{1}{4}+\frac{1}{2}$$ = $$$$
$$\frac{3}{4}$$ $$$$
Pati didžiausia reikšmė $$\frac{3}{4}$$
Atsakymas: $$\frac{3}{4}$$