A Radical Functional Equation | Algebra

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  • เผยแพร่เมื่อ 4 ก.ค. 2024
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ความคิดเห็น • 13

  • @cicik57
    @cicik57 9 วันที่ผ่านมา +1

    substitute f(t) and solve:
    √x²+1 = (t+1)x > 0
    x²+1 = (t+1)²x²
    ((t+1)²-1)x² =1
    x² = 1/(t²+2t)
    so f(t)= 1/(t²+2t)

  • @txikitofandango
    @txikitofandango 9 วันที่ผ่านมา

    How I did it. Assume x>0, worry about negative x later.
    Multiply input of x by its conjugate, top and bottom. Change x to sqrt(x). This gives
    f(1/(sqrt(x^2 + x) + x)) = x.
    inverse of both sides: 1/(sqrt(x^2 + x) + x = f^-1(x) = y.
    SWAP x y
    1/(sqrt(y^2 + y) + y = f^-1(y) = x
    Solve for y = f(x).
    sqrt(y^2 + y) = 1/x - y
    y^2 + y = 1/x^2 - 2y/x + y^2
    The y^2 cancel out!
    y = 1/(x^2(1+2/x) = 1/(x^2 + 2x) = f(x)
    Your way was easier

  • @MrGeorge1896
    @MrGeorge1896 8 วันที่ผ่านมา +1

    Nice problem and easy so solve. 😊
    btw: at 7:04 you wrote root x² - 1 instead of root x² + 1

  • @dwittn9193
    @dwittn9193 9 วันที่ผ่านมา

    Nice

  • @DonEnsley-mathdrum
    @DonEnsley-mathdrum 10 วันที่ผ่านมา +3

    f(x) = 1/(x² +2x)

    • @d-8664
      @d-8664 10 วันที่ผ่านมา +1

      x cannot be 0 or -2.

    • @tunistick8044
      @tunistick8044 10 วันที่ผ่านมา

      ​@@d-8664 I want it to can

    • @HoSza1
      @HoSza1 9 วันที่ผ่านมา

      It was in the video.

    • @DonEnsley-mathdrum
      @DonEnsley-mathdrum 9 วันที่ผ่านมา

      That's right! X cannot be 0 or -2! That's because it makes the bottom part 0 which is a no no. Domain is x not 0 and x not -2

  • @YouTube_username_not_found
    @YouTube_username_not_found 10 วันที่ผ่านมา

    Goal: find all functions f: ℝ→ℝ such that for all x∈ℝ* ; f([√(x^2+1) − x]/x) = x^2
    One first notices that x = √(x^2) if x>0 and that x = −√(x^2) if x0 , which means the equation is equivalent to the following:
    For all t ∈ ℝ>0 ; f([√(t+1) − √t]/√t) = t and f([√(t+1) + √t]/−√t) = t. Let's solve the 1st part.
    Next, let u = [√(t+1) − √t]/√t ie. u√t = √(t+1) − √t ie. (u+1)√t = √(t+1) ie. (u+1)^2×t = t+1 and t+1≥0 (which is verified because t ∈ ℝ>0) ie. [(u+1)^2−1]×t = 1 ie. t = 1/[(u+1)^2−1] = 1/[u(u+2)].
    For all t ∈ ℝ>0 ; t+1>t which implies √(t+1)>√t ⇒ √(t+1) −√t >0 and 1/√t >0 ⇒ u = [√(t+1) − √t]/√t ∈ℝ>0 so the 1st part becomes: For all u ∈ ℝ>0 ; f(u) = 1/[u(u+2)]
    Similarly, let v = [√(t+1) + √t]/−√t ie. v√t = −√(t+1) − √t ie. (v+1)√t = −√(t+1) ie. (v+1)^2×t = t+1 and t+1≥0 ... same work as above ... t = 1/[v(v+2)].
    For all t ∈ ℝ>0 ; t+1>1 and t>0 which implies √(t+1)>1 and √t>0 ⇒ √(t+1) +√t >1 and 1/√t >0 ⇒ [√(t+1) + √t]/√t >0 ⇒ v = [√(t+1) + √t]/−√t ∈ℝ

    • @YouTube_username_not_found
      @YouTube_username_not_found 10 วันที่ผ่านมา

      I realized I made mistakes but I am too tired 😩to correct them.

  • @phill3986
    @phill3986 11 วันที่ผ่านมา

    😊😊😊👍👍👍

  • @barakathaider6333
    @barakathaider6333 10 วันที่ผ่านมา

    👍