Question Number 101156 by I want to learn more last updated on 30/Jun/20

Find the range of the function:       h : x   =  2  −  x^2  sin(x),      x  ≥  0

Answered by 1549442205 last updated on 07/Jul/20

x^2 sinx+x−2=0.we look at it like as a  quadratic equation with respect x.Then  Δ=(−1)^2 −4sinx.(−2)=1+8sinx  The equation always has roots,so  Δ≥0⇒1+8sinx≥0⇔sinx≤−(1/8)=sin^(−1) (−(1/8))  ⇔       𝛑−arcsin(−(1/8))+2m𝛑≤x≤2𝛑+arcsin(−(1/8))+2m𝛑(m∈N^∗ )

Commented by1549442205 last updated on 01/Jul/20

 If so then there is also a solution for it,  see above.The figure below shows that   all the amgles have one side which is the ray   AG and the second side  in the blue part  belongs to the range of x(sinϕ=−(1/8),−(π/2)<ϕ<0)

Commented byI want to learn more last updated on 01/Jul/20

sir, i thought it should be like.  y  =  2 − x^2 sin(x),   but i cannot proceed

Commented by1549442205 last updated on 01/Jul/20

You are welcome sir.

Commented byI want to learn more last updated on 01/Jul/20

Thanks sir