$\displaystyle f(x)=2^{x+1}-2^{-x}$. If $\displaystyle g(x)=f^{-1}(x)$ then what is the value of $\displaystyle g'(1)$? Im telling my friend my she made a mistake...but she won't belive me. I just need a 2nd opinion
thanks!
Edit: Sorry, this method doesn't work. Kudos to kalagota for figuring out why.
g'(1) = 1/f'(1)
f'(x) = $\displaystyle 2log(2)2^x+log(2)2^{-x}$
f'(1) = 4log2+.5log(2)=4.5log(2)
g'(1) = $\displaystyle \frac{1}{4.5log(2)}$
There are other ways of expressing this answer, so if you got something different you are not necessarily wrong (assuming I'm right that is).
if g(x) is said to be the inverse of f(x), then it is not necessary that $\displaystyle g(x) = \frac{1}{f(x)}$ (have you tried graphing them?)
do not confuse the inverse $\displaystyle f^{-1}(x)$ with the power $\displaystyle (f(x))^{-1} = \frac{1}{f(x)}$ they are different..
g = $\displaystyle f^{-1}$
therefore g(f(x)) = x
let f(x) = y
g(y) = x
differentiating both sides with respect to x and using the chain rule
$\displaystyle
\frac {dy}{dx} \frac{dg}{dy}$ = 1
$\displaystyle \frac{dg}{dy}$ = $\displaystyle \frac{1}{\frac{dy}{dx}}$
so the derivative of the inverse is the reciprocal of the derivative or I have made a mistake somewhere.
well, i'll do it this way..
if $\displaystyle y = 2^{x+1}-2^{-x}$, then the inverse is of the form $\displaystyle x = 2^{y+1}-2^{-y}$ (that is interchange x and y then solve for y in terms of x.. but i'll not do it..)
if we take the derivative of the latter equation (the inverse) implicitly, we would get
$\displaystyle 1 = 2^{y+1} \ln 2 \cdot \frac{dy}{dx} - 2^{-y} \ln 2 \cdot \left( - \frac{dy}{dx} \right) = 2^{y+1} \ln 2 \cdot \frac{dy}{dx} + 2^{-y} \ln 2 \cdot \left( \frac{dy}{dx} \right)$
factoring and isolating $\displaystyle \frac{dy}{dx}$, we would have
$\displaystyle \frac{dy}{dx} = \dfrac{1}{2^{y+1} \ln 2 + 2^{-y} \ln 2} = \dfrac {1}{\ln 2} \cdot \dfrac{1}{2^{y+1} + 2^{-y}}$
so from $\displaystyle x = 2^{y+1}-2^{-y}$, if x=1, then
$\displaystyle 1 = 2^{y+1}-2^{-y} = \frac{2^{2y+1} -1}{2^y} \implies 2^y = 2\cdot 2^{2y} - 1$, so if we set $\displaystyle u=2^y$, we would get a quadratic equation of the form $\displaystyle 2u^2 - u - 1 =0$ which factors $\displaystyle (2u + 1)(u - 1) = 0$ or u = 1 or $\displaystyle u = -\frac{1}{2} $; but the second one cannot occur since exponential is always positive.. therefore, u = 1 implying y = 0..
going back to $\displaystyle \frac{dy}{dx} = \dfrac {1}{\ln 2} \cdot \dfrac{1}{2^{y+1} + 2^{-y}}$ evaluated at y = 0, we have
$\displaystyle
\dfrac {1}{\ln 2} \cdot \dfrac{1}{2^{1} + 2^{0}} = \frac{1}{3 \ln 2} \approx 0.480898
$
too long solution but it is correct..
look at the attachment..
the pink one is the fucntion f(x)..
solid red is the inverse g(x)
dashed red is the tangent line at x=0
the blue-green is the identity function y=x..
ohh..
pardon me.. i actually found where was your mistake..
yes it is true that g(f(x)) = x
if you set y=f(x), then g(y)=x..
the mistake was when you took the derivative with respect to x.. (when you used your chain rule..)
that should be, $\displaystyle \frac{d}{dx}\left( g(y) \right) \cdot \frac{dy}{dx} = 1$
and it will give you $\displaystyle \frac{dy}{dx} = f'(x) = 1$?
maybe this is a nicer approach..
f(g(x)) = x
thus, $\displaystyle f'(g(x)) \cdot g'(x) = 1 \implies g'(x) = \frac{1}{f'(g(x))}$..
or g(f(x)) = x
and so, $\displaystyle g'(f(x)) \cdot f'(x) = 1 $
in both cases, you would have to find what $\displaystyle g(x) = f^{-1}(x)$ looks like.. (then, not so nice again.. Ü)