A agl89 New member Joined Jan 28, 2014 Messages 4 Jan 29, 2014 #1 Finding f(^-1)'(-1) Find f(^-1)'(-1) for f(x) = 2x / (x - 3). d/dx f(^-1)(-1) = 1 / (f'(f^-1(-1))) 2x / (x - 3) = -1 [2x / (x - 3)] + 1 = 0 ...I'm not sure where to go from here. Last edited: Jan 29, 2014
Finding f(^-1)'(-1) Find f(^-1)'(-1) for f(x) = 2x / (x - 3). d/dx f(^-1)(-1) = 1 / (f'(f^-1(-1))) 2x / (x - 3) = -1 [2x / (x - 3)] + 1 = 0 ...I'm not sure where to go from here.
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Jan 29, 2014 #2 agl89 said: Find f(^-1)'(-1) for f(x) = 2x / (x - 3). d/dx f(^-1)(-1) = 1 / (f'(f^-1(-1))) 2x / (x - 3) = -1 [2x / (x - 3)] + 1 = 0 ...I'm not sure where to go from here. Click to expand... If we solve that we get \(\displaystyle x=1\). So \(\displaystyle {f^{ - 1}}\left( { - 1} \right) = 1\)
agl89 said: Find f(^-1)'(-1) for f(x) = 2x / (x - 3). d/dx f(^-1)(-1) = 1 / (f'(f^-1(-1))) 2x / (x - 3) = -1 [2x / (x - 3)] + 1 = 0 ...I'm not sure where to go from here. Click to expand... If we solve that we get \(\displaystyle x=1\). So \(\displaystyle {f^{ - 1}}\left( { - 1} \right) = 1\)
A agl89 New member Joined Jan 28, 2014 Messages 4 Jan 29, 2014 #3 pka said: If we solve that we get \(\displaystyle x=1\). So \(\displaystyle {f^{ - 1}}\left( { - 1} \right) = 1\) Click to expand... I had trouble solving for x, but I figured it out... So, x = 1 => f(1) = -1 => f^-1(-1)= 1 ∴ d/dx f^-1(-1) = 1 / f'(1) d/dx [2x / (x - 3)] = -6 / (x - 3)^2 f'(1) = -6 / (1 - 3)^2 = -3/2 ∴ d/dx f^-1(-1) = 1 / (-3/2) = -2/3 Is this correct?
pka said: If we solve that we get \(\displaystyle x=1\). So \(\displaystyle {f^{ - 1}}\left( { - 1} \right) = 1\) Click to expand... I had trouble solving for x, but I figured it out... So, x = 1 => f(1) = -1 => f^-1(-1)= 1 ∴ d/dx f^-1(-1) = 1 / f'(1) d/dx [2x / (x - 3)] = -6 / (x - 3)^2 f'(1) = -6 / (1 - 3)^2 = -3/2 ∴ d/dx f^-1(-1) = 1 / (-3/2) = -2/3 Is this correct?
