Answer to Question #219211 in Calculus for Sarita bartwal

Question

Find the range of 1.f(x,y,z)= z/(x^2-y^2)
2. f(x,y)= x sin(1/x)+ y sin(1/y)
3. f(x,y,z)= 1/(√ (4-x^2-y^2-z^2)

Answer 1

Answer:-

1.

$$x^2-y^2\not=0$$

If $z\geq 0, x^2>y^2,$ then $f(x, y,z)\geq0.$

If $z\geq 0, x^2<y^2,$ then $f(x, y,z)\leq0.$

If $z\leq 0, x^2>y^2,$ then $f(x, y,z)\leq0.$

If $z\leq 0, x^2<y^2,$ then $f(x, y,z)\geq0.$

Range: $(-\infin, \infin)$

2. Let $u(x)=x\sin(\dfrac{1}{x})$

If $x\to\pm \infin,$ then $u(x)\to 1^{-}$

$$u'=\sin(\dfrac{1}{x})-\dfrac{1}{x}\cos(\dfrac{1}{x})$$ $$u'=0=>\sin(\dfrac{1}{x})-\dfrac{1}{x}\cos(\dfrac{1}{x})=0$$ $$x_1=-0.22255, x_2=0.22255$$ $$u(-0.22255)=u(0.22255)\approx-0.21723$$ $$-0.21723\leq u<1, x\not=0$$ $$-0.21723-0.21723\approx-0.4345$$

Range: $[-0.4345, 2)$

3.

$$4-x^2-y^2-z^2>0$$ $$0\leq x^2+y^2+z^2<4$$

Then

$$0<\sqrt{4-x^2-y^2-z^2}\leq2$$

Range: $\bigg[\dfrac{1}{2}, \infin\bigg)$