Răspuns:
f(x)=x³-3x+m
f `(x)=3x²-3
f `(x)=0
3x²-3=0
x1= -1
x2=1
f(-1)
TAbel
m l-∞ -1 1 +∞
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m-1l- - - - - - 0+ + +
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m+1l- - - -0+ + ++ +
Calculezi f(-1)=- 1³-3(-1)+m=-1+3+m=m+2
f(1)=1³-3*1+m=1-3+m=m-2
Limitele in caapete
x->-∞lim fx)=lim(x³-3x+m)= -∞
x->+∞limf(x)=lim(x³+3x+m)=+∞
Tabel radacini ai intre schimbari de semn+/- sau invers si in punctele unde se anuleaza derivata
X l-∞ (-1) +1 +∞
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\f(x) l- m+2 m-2 +
m l
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m<-2l- - - - +(ai solituie α∈ (1 si +∞)
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m=-2 l- ----------------------0--------------------- --- -----++ (solutie dubla in x= -1 si α3∈ (1,+ ∞)
-2<m<2l- - - -0+ + +++++- -------------+ (α∈(-∞,-1) α2∈(-1,1) α3>1
m=2 l--------------------------++++++++++++++0+++++++++++α1,∈(-∞ -1) α2=α3
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m>2 l- -------------------------------------+++++++++α∈(1∞,)
daca ai neclaritati intreaba-ma
Explicație pas cu pas:
