Răspuns:
f `(x)=(lnsin5x)) `
Aplici formula (ln u) `=u `/u unde u= sin 5x
f `(x)=(sin5x) `/sin5x=5cos5x/sin5x=5 ctgx
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Calculezi derivata si rezolvi ecuatia f `(x)=0
f `(x)=√3*2x-√3=2√3x-√3
f `(x)=2√3x-√3=0 2x-1=0 x=1/2
Pt x<1/2 2x-1<0 => f `(x)<0
Pt x>1/2 2x-1>0 f `(x)>0
Derivata isi schimba semnul de-o parte si alta a lui x=1/2 =>x=1/2 punct de extrem. f(1/2)=√3*1/4-√3*1/2+3
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F `(x)=3*5*x^4-2*3*x²=15x^4-6x²=3x²(5x²-2)
f `(x)=3x²(5x²-2)=0 x²=0 ;5x²=2 =>
x1=x2=0 x3=-√(2/5) x4=√(2/5)Tabelul de semne
√2/5=√0,5≈0,1
x l-∞.......................-√5/4....0..√5/4.................+∞
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x² l+ + + + + + + + + +
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5x²-2 l+ + + + 0 - - -0 + + + +
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x²(5x²-2) l+ + + + 0-----0-- - -o + + + + + +
f `(x) isi schimba semnul de-o parte si de alta a lui -√(2/5) si √(2/5)=>
acestea sunt puncte de extrem
f(√2/5)=√3*(2/5)^4*√(2/5)-√3*√(2/5)=-√(6/5)**(21/25
In mod analog calculezi pe f( -√(2/5))
Aplici formula (ln u) `=u `/u unde u= sin 5x
f `(x)=(sin5x) `/sin5x=5cos5x/sin5x=5 ctgx
________________________
Calculezi derivata si rezolvi ecuatia f `(x)=0
f `(x)=√3*2x-√3=2√3x-√3
f `(x)=2√3x-√3=0 2x-1=0 x=1/2
Pt x<1/2 2x-1<0 => f `(x)<0
Pt x>1/2 2x-1>0 f `(x)>0
Derivata isi schimba semnul de-o parte si alta a lui x=1/2 =>x=1/2 punct de extrem. f(1/2)=√3*1/4-√3*1/2+3
___________________--
F `(x)=3*5*x^4-2*3*x²=15x^4-6x²=3x²(5x²-2)
f `(x)=3x²(5x²-2)=0 x²=0 ;5x²=2 =>
x1=x2=0 x3=-√(2/5) x4=√(2/5)Tabelul de semne
√2/5=√0,5≈0,1
x l-∞.......................-√5/4....0..√5/4.................+∞
_______________________________
x² l+ + + + + + + + + +
__________________________________
5x²-2 l+ + + + 0 - - -0 + + + +
______________________________________
x²(5x²-2) l+ + + + 0-----0-- - -o + + + + + +
f `(x) isi schimba semnul de-o parte si de alta a lui -√(2/5) si √(2/5)=>
acestea sunt puncte de extrem
f(√2/5)=√3*(2/5)^4*√(2/5)-√3*√(2/5)=-√(6/5)**(21/25
In mod analog calculezi pe f( -√(2/5))
Explicație pas cu pas:
