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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

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Step-by-step solution

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

Domain:

Square root requires non-negative argument.

x≥0x\ge 0

Domain: [0,∞)[0,\infty)

Range:

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

    y=xy=\sqrt{x}
  2. Square both sides.

    y2=xy^2=x
  3. y≥0y\ge 0 (principal square root).

  4. x=y2≥0x=y^2\ge 0, which is in domain for all y≥0y\ge 0.

    As yy goes from 0 to ∞\infty, xx goes from 0 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:

  5. Let y=g(x)y=g(x).

    y=2xy=2^{x}
  6. y>0y>0 always since 2x>02^x>0.

  7. x=log⁡2yx=\log_2 y, defined for y>0y>0.

    As x→−∞x\to -\infty, y→0+y\to 0^+; as x→∞x\to \infty, y→∞y\to \infty.

    Range: (0,∞)(0,\infty)

    (0,\infty)

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

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

    Range:

  8. Let y=m(x)y=m(x).

    y=x2y=x^2
  9. x=±yx=\pm \sqrt{y}.

  10. Real xx requires y≥0y\ge 0.

    For all y≥0y\ge 0, real xx exists.

    Minimum y=0y=0 at x=0x=0; y→∞y\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:

  11. Let y=n(x)y=n(x).

    y=x3y=x^3
  12. x=y3x=\sqrt[3]{y}.

  13. Cube root defined for all real yy.

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

    Range: (−∞,∞)(-\infty,\infty)

    (-\infty,\infty)

    3. u(x)=x2−4u(x)=\sqrt{x^2-4}

    Domain:

    Inside square root ≥0\ge 0.

    x2−4≥0x^2-4\ge 0 x2≥4x^2\ge 4

    x≤−2x\le -2 or x≥2x\ge 2

    Range:

  14. Let y=u(x)y=u(x).

    y=x2−4y=\sqrt{x^2-4} y≥0y\ge 0
  15. Square both sides.

    y2=x2−4y^2=x^2-4 x2=y2+4x^2=y^2+4
  16. ∣x∣=y2+4≥2|x|=\sqrt{y^2+4}\ge 2, since y2+4≥4y^2+4\ge 4.

    Always true for y≥0y\ge 0.

    Minimum y=0y=0 at ∣x∣=2|x|=2; y→∞y\to\infty as ∣x∣→∞|x|\to\infty.

    Range: [0,∞)[0,\infty)

    [0,\infty)

    4. v(x)=ln⁡(x2−1)v(x)=\ln(x^2-1)

    Domain:

    Argument >0>0.

    x2−1>0x^2-1>0 ∣x∣>1|x|>1

    Range:

  17. Let y=v(x)y=v(x).

    y=ln⁡(x2−1)y=\ln(x^2-1)
  18. x2−1=eyx^2-1=e^y

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

  19. Always true for all real yy.

    As ∣x∣→1+|x|\to 1^+, x2−1→0+x^2-1\to 0^+, y→−∞y\to -\infty.

    As ∣x∣→∞|x|\to\infty, y→∞y\to\infty.

    Range: (−∞,∞)(-\infty,\infty)

    (-\infty,\infty)

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