f(x)={x^{2}}/3+x-{x}/3
Clearly it has two real solutions i.e. 0 and -2 suppose a=0,-2
f(x,a)=a^3 +4x^2 -x=0
f(-2,-2)=10 which does NOT satisfy the given condition.
f(0,-2)=-8 which does NOT satisfy the given condition.

f(-2,0)=18 which does NOT satisfy the given condition.

Please help me,. (prove : a is a only solution of function f(x)=(x)^2÷3 + x + (-x)1÷3 on -inf;-1] then a^3 +4x^2 -x=0

Reply

f(x)={x^{2}}/3+x-{x}/3 Clearly it has two real solutions i.e. 0 and -2 suppose a=0,-2 f(x,a)=a^3 +4x^2 -x=0 f(-2,-2)=10 which does NOT satisfy the given condition. f(0,-2)=-8 which does NOT satisfy the given condition.

f(-2,0)=18 which does NOT satisfy the given condition.

f(0,0)=0 which satisfies the given condition.