Question

Fie f(-pi/2, pi/2) ->  R, f(x)= sinx + cos x. Imaginea functiei f este ?

Answer 1

$f(x)=sinx+cosx=\sqrt{2}(sinx \cdot cos\frac{\pi}{4}+sin\frac{\pi}{4} \cdot cosx)=\sqrt{2} \cdot sin(x+\frac{\pi}{4})$
-π/4<x+π/4<3π/4
Daca (x+π/4)∈[0,π/2] si functia sinus crescatoare=>
sin0≤sin(x+π/4)≤sinπ/2|*√2=>0≤f(x)≤√2

Daca (x+π/4)∈(π/2,3π/4) si functia sinus descrescatoare, folosind ideea de mai sus obtinem = >sinπ/2≤sin(x+π/4)≤sin3π/4|*√2=>1<f(x)<√2.

Daca (x+π/4)∈(-π/4,0 si functia sinus crescatoare, folosind ideea de mai sus obtinem = >sin(-π/4)≤sin(x+π/4)≤sin0|*√2=>-1<f(x)<0.
In concluzie daca x∈(-π/2;π/2) atunci -1<f(x)≤√2, adica Imf=(-1;√2].