Find-the-range-of-the-function-h-x-2-x-2-sin-x-x-0-

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} - 4 sinx . (- 2) = 1 + 8 sinx

The equation always has roots, so

Δ ⩾ 0 \Rightarrow 1 + 8 sinx ⩾ 0 \Leftrightarrow sinx ⩽ - \frac{1}{8} = \sin^{- 1} (- \frac{1}{8})

\Leftrightarrow π - \arcsin (- \frac{1}{8}) + 2 m π ⩽ x ⩽ 2 π + \arcsin (- \frac{1}{8}) + 2 m π (m \in N^{*})

Commented by 1549442205 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 (\sin φ = - \frac{1}{8}, - \frac{π}{2} < φ < 0)

Commented by I 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 by 1549442205 last updated on 01/Jul/20

You are welcome sir .

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

Thanks sir