Solve the inequalities: 1/(x+6) < 2/3-3n. Find domain and range of functions g(x)=x-2, h(x)=7n/(x2-f), k(x)=(x+1)/sqrt(2x+1). Find the range of values of x for which |2x-4| <= 4.

Mathematics
Solve the inequalities: 1/(x+6) < 2/3-3n. Find domain and range of functions g(x)=x-2, h(x)=7n/(x2-f), k(x)=(x+1)/sqrt(2x+1). Find the range of values of x for which |2x-4| <= 4.

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Answer

[0,\infty)

1(a) f(x)=xf(x)=\sqrt{x}

Domain:

Square root requires non-negative argument.

x0x\ge 0

Domain: [0,)[0,\infty)

Range:

Step 1: Let y=f(x)y=f(x).

y=xy=\sqrt{x}

Step 2: Square both sides.

y2=xy^2=x

Step 3: y0y\ge 0 (principal square root).

Step 4: x=y20x=y^2\ge 0, which is in domain for all y0y\ge 0.

As yy goes from 00 to \infty, xx goes from 00 to \infty.

Range: [0,)[0,\infty)

[0,\infty)

1(b) g(x)=2xg(x)=2^{x}

Domain:

Exponential defined for all real xx.

D=(,)D=(-\infty,\infty)

Range:

Step 1: Let y=g(x)y=g(x).

y=2xy=2^{x}

Step 2: y>0y>0 always since 2x>02^x>0.

Step 3: x=log2yx=\log_2 y, defined for y>0y>0.

As xx\to -\infty, y0+y\to 0^+; as xx\to \infty, yy\to \infty.

Range: (0,)(0,\infty)

(0,\infty)

2(a) m(x)=x2m(x)=x^2

Domain: R\mathbb{R} (polynomial).

Range:

Step 1: Let y=m(x)y=m(x).

y=x2y=x^2

Step 2: x=±yx=\pm \sqrt{y}.

Step 3: Real xx requires y0y\ge 0.

For all y0y\ge 0, real xx exists.

Minimum y=0y=0 at x=0x=0; yy\to\infty as x|x|\to\infty.

Range: [0,)[0,\infty)

[0,\infty)

2(b) n(x)=x3n(x)=x^3

Domain: R\mathbb{R}.

Range:

Step 1: Let y=n(x)y=n(x).

y=x3y=x^3

Step 2: x=y3x=\sqrt[3]{y}.

Step 3: Cube root defined for all real yy.

Bijective from R\mathbb{R} to R\mathbb{R}.

Range: (,)(-\infty,\infty)

(-\infty,\infty)

3. u(x)=x24u(x)=\sqrt{x^2-4}

Domain:

Inside square root 0\ge 0.

x240x^2-4\ge 0

x24x^2\ge 4

x2x\le -2 or x2x\ge 2

Range:

Step 1: Let y=u(x)y=u(x).

y=x24y=\sqrt{x^2-4}

y0y\ge 0

Step 2: Square both sides.

y2=x24y^2=x^2-4

x2=y2+4x^2=y^2+4

Step 3: x=y2+42|x|=\sqrt{y^2+4}\ge 2, since y2+44y^2+4\ge 4.

Always true for y0y\ge 0.

Minimum y=0y=0 at x=2|x|=2; yy\to\infty as x|x|\to\infty.

Range: [0,)[0,\infty)

[0,\infty)

4. v(x)=ln(x21)v(x)=\ln(x^2-1)

Domain:

Argument >0>0.

x21>0x^2-1>0

x>1|x|>1

Range:

Step 1: Let y=v(x)y=v(x).

y=ln(x21)y=\ln(x^2-1)

Step 2: x21=eyx^2-1=e^y

x2=ey+1>1x^2=e^y+1>1 (since ey>0e^y>0)

Step 3: Always true for all real yy.

As x1+|x|\to 1^+, x210+x^2-1\to 0^+, yy\to -\infty.

As x|x|\to\infty, yy\to\infty.

Range: (,)(-\infty,\infty)

(-\infty,\infty)

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Quick Answer

1(a) f(x)=sqrt(x) Domain: Square root requires non-negative argument. x 0 Domain: [0,) Range: Step 1: Let y=f(x).

Solve the inequalities: 1/(x+6) < 2/3-3n. Find domain and range of functions g(x)=x-2, h(x)=7n/(x2-f), k(x)=(x+1)/sqrt(2x+1). Find the range of values of x for which |2x-4| <= 4.
Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

ScanSolve AI Answer
1(a) f(x)=sqrt(x) Domain: Square root requires non-negative argument. x 0 Domain: [0,) Range: Step 1: Let y=f(x). y=sqrt(x) Step 2: Square both sides. y^2=x Step 3: y 0 (principal square root). Step 4: x=y^2 0, which is in domain for all y 0. As y goes from 0 to , x goes from 0 to . Range: [0,) [0,) 1(b) g(x)=2^x Domain: Exponential defined for all real x. D=(-,) Range: Step 1: Let y=g(x). y=2^x Step 2: y>0 always since 2^x>0. Step 3: x=_2 y, defined for y>0. As x -, y 0^+; as x , y . Range: (0,) (0,) 2(a) m(x)=x^2 Domain: R (polynomial). Range: Step 1: Let y=m(x). y=x^2 Step 2: x=± sqrt(y). Step 3: Real x requires y 0. For all y 0, real x exists. Minimum y=0 at x=0; y as |x|. Range: [0,) [0,) 2(b) n(x)=x^3 Domain: R. Range: Step 1: Let y=n(x). y=x^3 Step 2: x=[3]y. Step 3: Cube root defined for all real y. Bijective from R to R. Range: (-,) (-,) 3. u(x)=sqrt(x^2-4) Domain: Inside square root 0. x^2-4 0 x^2 4 x -2 or x 2 Range: Step 1: Let y=u(x). y=sqrt(x^2-4) y 0 Step 2: Square both sides. y^2=x^2-4 x^2=y^2+4 Step 3: |x|=sqrt(y^2+4) 2, since y^2+4 4. Always true for y 0. Minimum y=0 at |x|=2; y as |x|. Range: [0,) [0,) 4. v(x)=(x^2-1) Domain: Argument >0. x^2-1>0 |x|>1 Range: Step 1: Let y=v(x). y=(x^2-1) Step 2: x^2-1=e^y x^2=e^y+1>1 (since e^y>0) Step 3: Always true for all real y. As |x| 1^+, x^2-1 0^+, y -. As |x|, y. Range: (-,) (-,)