28 uždavinys

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Raskite didžiausią funkcijos f(x) = 12⋅cos(2⋅x)+sin(x)\frac{1}{2}\cdot cos(2\cdot x)+sin(x) reikšmę intervale [0; π2\frac{\pi}{2}]

Sprendimas.

Randame funkcijos f(x) išvestinę.

 ( 
 1
/ 2
* cos( 2* x)+sin(x))′  = 
 ( 
 1
/ 2
* cos( 2* x)+sin(x))′ = (12⋅cos(2⋅x)+sin(x))′(\frac{1}{2}\cdot cos(2\cdot x)+sin(x))' = 
(12⋅cos(2⋅x)+sin(x))′{\normalsize (\frac{1}{2}\cdot cos(2\cdot x)+sin(x))'} = (12⋅cos(2⋅x))′+sin(x)′{\normalsize (\frac{1}{2}\cdot cos(2\cdot x))'+sin(x)'}
Paaiškinimas:
Sumos išvestinė (f+g)′ = f′ + g′
 ( 
 1
/ 2
* cos( 2* x))′+ sin(x)′ = (12⋅cos(2⋅x))′+sin(x)′(\frac{1}{2}\cdot cos(2\cdot x))'+sin(x)' = 
12⋅cos(2⋅x){\normalsize \frac{1}{2}\cdot cos(2\cdot x)} = 12⋅cos(2⋅x)′{\normalsize \frac{1}{2}\cdot cos(2\cdot x)'}
Paaiškinimas:
1/2 iškeltas prieš skliaustus
 
 1
/ 2
* cos( 2* x)′+ sin(x)′ = 12⋅cos(2⋅x)′+sin(x)′\frac{1}{2}\cdot cos(2\cdot x)'+sin(x)' = 
cos(2⋅x)′{\normalsize cos(2\cdot x)'} = sin(2⋅x)⋅(2⋅x)′{\normalsize sin(2\cdot x)\cdot (2\cdot x)'}
Paaiškinimas:
cos(x) išvestinė yra sin(x)
Kompozicijos g(f(x)) išvestinė yra g(y)*f(x) f(x) = (2*x)′
- 
 1
/ 2
* sin( 2* x)* ( 2* x)′+ sin(x)′ = −12⋅sin(2⋅x)⋅(2⋅x)′+sin(x)′-\frac{1}{2}\cdot sin(2\cdot x)\cdot (2\cdot x)'+sin(x)' = 
(2⋅x)′{\normalsize (2\cdot x)'} = 2{\normalsize 2}
Paaiškinimas:
x išvestinė yra 1
- 
 1
/ 2
* sin( 2* x)* 2+ sin(x)′ = −12⋅sin(2⋅x)⋅2+sin(x)′-\frac{1}{2}\cdot sin(2\cdot x)\cdot 2+sin(x)' = 
12⋅sin(2⋅x)⋅2{\normalsize \frac{1}{2}\cdot sin(2\cdot x)\cdot 2} = sin(2⋅x){\normalsize sin(2\cdot x)}
-sin( 2* x)+ sin(x)′ = −sin(2⋅x)+sin(x)′-sin(2\cdot x)+sin(x)' = 
sin(x)′{\normalsize sin(x)'} = cos(x){\normalsize cos(x)}
Paaiškinimas:
sin(x) išvestinė yra cos(x)
-sin( 2* x)+cos(x) = −sin(2⋅x)+cos(x)-sin(2\cdot x)+cos(x) = 
sin(2⋅x){\normalsize sin(2\cdot x)} = (2⋅sin(x)⋅cos(x)){\normalsize (2\cdot sin(x)\cdot cos(x))}
Paaiškinimas:
Sinuso argumento žeminimo formulė sin(2⋅a)=2⋅sin(a)⋅cos(a){\normalsize sin(2\cdot a) = 2\cdot sin(a)\cdot cos(a)}
-( 2* sin(x)* cos(x))+cos(x) = −(2⋅sin(x)⋅cos(x))+cos(x)-(2\cdot sin(x)\cdot cos(x))+cos(x) = 
−(2⋅sin(x)⋅cos(x)){\normalsize -(2\cdot sin(x)\cdot cos(x))} = −2⋅sin(x)⋅cos(x){\normalsize -2\cdot sin(x)\cdot cos(x)}
- 2* sin(x)* cos(x)+cos(x) = −2⋅sin(x)⋅cos(x)+cos(x)-2\cdot sin(x)\cdot cos(x)+cos(x) = 
−2⋅sin(x)⋅cos(x)+cos(x){\normalsize -2\cdot sin(x)\cdot cos(x)+cos(x)} = −cos(x)⋅(2⋅sin(x)−1){\normalsize -cos(x)\cdot (2\cdot sin(x)-1)}
Paaiškinimas:
cos(x) iškeltas prieš skliaustus
- cos(x)* ( 2* sin(x)-1)−cos(x)⋅(2⋅sin(x)−1)-cos(x)\cdot (2\cdot sin(x)-1)
(12⋅cos(2⋅x)+sin(x))′(\frac{1}{2}\cdot cos(2\cdot x)+sin(x))'  = 
12⋅cos(2⋅x)′+sin(x)′\frac{1}{2}\cdot cos(2\cdot x)'+sin(x)'  = 
−12⋅sin(2⋅x)⋅(2⋅x)′+sin(x)′-\frac{1}{2}\cdot sin(2\cdot x)\cdot (2\cdot x)'+sin(x)'  = 
−12⋅sin(2⋅x)⋅2+sin(x)′-\frac{1}{2}\cdot sin(2\cdot x)\cdot 2+sin(x)'  = 
−sin(2⋅x)+sin(x)′-sin(2\cdot x)+sin(x)'  = 
−sin(2⋅x)+cos(x)-sin(2\cdot x)+cos(x)  = 
−(2⋅sin(x)⋅cos(x))+cos(x)-(2\cdot sin(x)\cdot cos(x))+cos(x)  = 
−cos(x)⋅(2⋅sin(x)−1)-cos(x)\cdot (2\cdot sin(x)-1)

Norint rasti ekstremumus (didžiausias/mažiausias reikšmes), išvestinę −cos(x)⋅(2⋅sin(x)−1)-cos(x)\cdot (2\cdot sin(x)-1) reikia prilyginti nuliui.

Gauname cos(x)=0cos(x) = 0 (1)

ir 2⋅sin(x)−1=02\cdot sin(x)-1 = 0 (2)

Iš (1) lygties gauname x = π2\frac{\pi}{2}

Iš (2) lygties gauname sin(x)=12sin(x) = \frac{1}{2},

x = π6\frac{\pi}{6}

Apskaičiuojame funkcijos f(x) reikšmes rastuose ektremumo taškuose x = π2\frac{\pi}{2} ir x = π6\frac{\pi}{6} bei pradiniame intervalo taške x = 0.

Kai x = 0:

 
 1
/ 2
* cos( 2* x)+sin(x)  = 
 
 1
/ 2
* cos( 2* x)+sin(x) = 12⋅cos(2⋅x)+sin(x)\frac{1}{2}\cdot cos(2\cdot x)+sin(x) = 
Paaiškinimas:
Keitimas x{\normalsize x} = 00.
 
 1
/ 2
* cos( 2* 0)+sin(0) = 12⋅cos(2⋅0)+sin(0)\frac{1}{2}\cdot cos(2\cdot 0)+sin(0) = 
2⋅0{\normalsize 2\cdot 0} = 00
 
 1
/ 2
* cos(0)+sin(0) = 12⋅cos(0)+sin(0)\frac{1}{2}\cdot cos(0)+sin(0) = 
sin(0){\normalsize sin(0)} = 00
 
 1
/ 2
* cos(0)+0 = 12⋅cos(0)+0\frac{1}{2}\cdot cos(0)+0 = 
 
 1
/ 2
* cos(0) = 12⋅cos(0)\frac{1}{2}\cdot cos(0) = 
cos(0){\normalsize cos(0)} = 1{\normalsize 1}
 
 1
/ 2
* 1 = 12⋅1\frac{1}{2}\cdot 1 = 
12⋅1{\normalsize \frac{1}{2}\cdot 1} = 12{\normalsize \frac{1}{2}}
 
 1
/ 2
12\frac{1}{2}
12⋅cos(2⋅0)+sin(0)\frac{1}{2}\cdot cos(2\cdot 0)+sin(0)  = 
12⋅cos(0)\frac{1}{2}\cdot cos(0)  = 
12⋅1\frac{1}{2}\cdot 1  = 
12\frac{1}{2}

Kai x = π2\frac{\pi}{2}

 
 1
/ 2
* cos( 2* x)+sin(x)  = 
 
 1
/ 2
* cos( 2* x)+sin(x) = 12⋅cos(2⋅x)+sin(x)\frac{1}{2}\cdot cos(2\cdot x)+sin(x) = 
Paaiškinimas:
Keitimas x{\normalsize x} = π2{\normalsize \frac{\pi}{2}}.
 
 1
/ 2
* cos( 
 2* π
/ 2
)+sin( 
 π
/ 2
) = 12⋅cos(2⋅π2)+sin(π2)\frac{1}{2}\cdot cos(\frac{2\cdot \pi}{2})+sin(\frac{\pi}{2}) = 
2⋅π2{\normalsize \frac{2\cdot \pi}{2}} = π{\normalsize \pi}
 
 1
/ 2
* cos(π)+sin( 
 π
/ 2
) = 12⋅cos(π)+sin(π2)\frac{1}{2}\cdot cos(\pi)+sin(\frac{\pi}{2}) = 
cos(π){\normalsize cos(\pi)} = −1{\normalsize -1}
- 
 1
/ 2
* 1+sin( 
 π
/ 2
) = −12⋅1+sin(π2)-\frac{1}{2}\cdot 1+sin(\frac{\pi}{2}) = 
12⋅1{\normalsize \frac{1}{2}\cdot 1} = 12{\normalsize \frac{1}{2}}
- 
 1
/ 2
+sin( 
 π
/ 2
) = −12+sin(π2)-\frac{1}{2}+sin(\frac{\pi}{2}) = 
sin(π2){\normalsize sin(\frac{\pi}{2})} = 1{\normalsize 1}
- 
 1
/ 2
+1 = −12+1-\frac{1}{2}+1 = 
−12+1{\normalsize -\frac{1}{2}+1} = 12{\normalsize \frac{1}{2}}
 
 1
/ 2
12\frac{1}{2}
12⋅cos(2⋅π2)+sin(π2)\frac{1}{2}\cdot cos(\frac{2\cdot \pi}{2})+sin(\frac{\pi}{2})  = 
12⋅cos(π)+sin(π2)\frac{1}{2}\cdot cos(\pi)+sin(\frac{\pi}{2})  = 
−12+sin(π2)-\frac{1}{2}+sin(\frac{\pi}{2})  = 
−12+1-\frac{1}{2}+1  = 
12\frac{1}{2}

Kai x =  π6\frac{\pi}{6}

 
 1
/ 2
* cos( 2* x)+sin(x)  = 
 
 1
/ 2
* cos( 2* x)+sin(x) = 12⋅cos(2⋅x)+sin(x)\frac{1}{2}\cdot cos(2\cdot x)+sin(x) = 
Paaiškinimas:
Keitimas x{\normalsize x} = π6{\normalsize \frac{\pi}{6}}.
 
 1
/ 2
* cos( 
 2* π
/ 6
)+sin( 
 π
/ 6
) = 12⋅cos(2⋅π6)+sin(π6)\frac{1}{2}\cdot cos(\frac{2\cdot \pi}{6})+sin(\frac{\pi}{6}) = 
2⋅π6{\normalsize \frac{2\cdot \pi}{6}} = π3{\normalsize \frac{\pi}{3}}
 
 1
/ 2
* cos( 
 π
/ 3
)+sin( 
 π
/ 6
) = 12⋅cos(π3)+sin(π6)\frac{1}{2}\cdot cos(\frac{\pi}{3})+sin(\frac{\pi}{6}) = 
cos(π3){\normalsize cos(\frac{\pi}{3})} = 12{\normalsize \frac{1}{2}}
 
 1
/ 2
* 
 1
/ 2
+sin( 
 π
/ 6
) = 12⋅12+sin(π6)\frac{1}{2}\cdot \frac{1}{2}+sin(\frac{\pi}{6}) = 
12⋅12{\normalsize \frac{1}{2}\cdot \frac{1}{2}} = 14{\normalsize \frac{1}{4}}
 
 1
/ 4
+sin( 
 π
/ 6
) = 14+sin(π6)\frac{1}{4}+sin(\frac{\pi}{6}) = 
sin(π6){\normalsize sin(\frac{\pi}{6})} = 12{\normalsize \frac{1}{2}}
 
 1
/ 4
+ 
 1
/ 2
 = 14+12\frac{1}{4}+\frac{1}{2} = 
14+12{\normalsize \frac{1}{4}+\frac{1}{2}} = 34{\normalsize \frac{3}{4}}
 
 3
/ 4
34\frac{3}{4}
12⋅cos(2⋅π6)+sin(π6)\frac{1}{2}\cdot cos(\frac{2\cdot \pi}{6})+sin(\frac{\pi}{6})  = 
12⋅cos(π3)+sin(π6)\frac{1}{2}\cdot cos(\frac{\pi}{3})+sin(\frac{\pi}{6})  = 
12⋅12+sin(π6)\frac{1}{2}\cdot \frac{1}{2}+sin(\frac{\pi}{6})  = 
14+sin(π6)\frac{1}{4}+sin(\frac{\pi}{6})  = 
14+12\frac{1}{4}+\frac{1}{2}  = 
34\frac{3}{4}

Pati didžiausia reikšmė 34\frac{3}{4}

Atsakymas: 34\frac{3}{4}

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