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)

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

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

f(x) = 1/(x² +2x)

@d-8664

15 күн бұрын

x cannot be 0 or -2.

@tunistick8044

15 күн бұрын

​@@d-8664 I want it to can

@HoSza1

14 күн бұрын

It was in the video.

@DonEnsley-mathdrum

14 күн бұрын

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

Nice

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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

15 күн бұрын

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

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