Question Number 102328 by bemath last updated on 08/Jul/20

Answered by bobhans last updated on 08/Jul/20

(ii) y= 2ln(((√(8x−4))/x)) ⇒x ≥ (1/2)  y = ln(((8x−4)/x^2 )) ⇒e^y .x^2  = 8x−4  e^y .x^2 −8x+4 = 0; x = ((8 ± (√(64−16e^y )))/(2.e^y ))  x = ((4 ± 2(√(4−e^y )))/e^y ) ⇒f^(−1) (x) = ((4 ±(√(4−e^x )))/e^x )  for 4−e^x  ≥0 ⇒e^x  ≤ 4 ; x ≤ ln(4) ∧x≥(1/2)  (1/2)≤x≤ ln(4)

Commented byI want to learn more last updated on 08/Jul/20

Thanks sir