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Question Number 101172 by I want to learn more last updated on 01/Jul/20
Find the range of the function:        y  =  2  −  x^2  sin(x),      x ≥ 0
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}: \\ $$$$\:\:\:\:\:\:\mathrm{y}\:\:=\:\:\mathrm{2}\:\:−\:\:\mathrm{x}^{\mathrm{2}} \:\mathrm{sin}\left(\mathrm{x}\right),\:\:\:\:\:\:\mathrm{x}\:\geqslant\:\mathrm{0} \\ $$
Answered by 1549442205 last updated on 07/Jul/20
Since −1≤sinx≤1,so:−x^2 ≤x^2 sinx≤x^2   we infer:2−x^2 ≤ 2−x^2 sinx≤2+x^2   ⇒−∞<y=2−x^2 sinx<+∞  The value domain of y:(−∞;+∞)
$$\mathrm{Since}\:−\mathrm{1}\leqslant\mathrm{sinx}\leqslant\mathrm{1},\mathrm{so}:−\mathrm{x}^{\mathrm{2}} \leqslant\mathrm{x}^{\mathrm{2}} \mathrm{sinx}\leqslant\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{we}\:\mathrm{infer}:\mathrm{2}−\mathrm{x}^{\mathrm{2}} \leqslant\:\mathrm{2}−\mathrm{x}^{\mathrm{2}} \mathrm{sinx}\leqslant\mathrm{2}+\mathrm{x}^{\mathrm{2}} \\ $$$$\Rightarrow−\infty<\boldsymbol{\mathrm{y}}=\mathrm{2}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{sinx}}<+\infty \\ $$$$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{domain}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{y}}:\left(−\infty;+\infty\right) \\ $$
Commented by I want to learn more last updated on 01/Jul/20
Sir, is this the same thing as:  Range is all real  numbers.
$$\mathrm{Sir},\:\mathrm{is}\:\mathrm{this}\:\mathrm{the}\:\mathrm{same}\:\mathrm{thing}\:\mathrm{as}: \\ $$$$\mathrm{Range}\:\mathrm{is}\:\mathrm{all}\:\mathrm{real}\:\:\mathrm{numbers}. \\ $$
Commented by 1549442205 last updated on 01/Jul/20
the value domain of a function and   the range of a function are same concept
$$\mathrm{the}\:\mathrm{value}\:\mathrm{domain}\:\mathrm{of}\:\mathrm{a}\:\mathrm{function}\:\mathrm{and}\: \\ $$$$\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{a}\:\mathrm{function}\:\mathrm{are}\:\mathrm{same}\:\mathrm{concept} \\ $$

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