1. ## inverse

dont know how to start this:

$f(x) = 5 + 5 x + 4 e^x$

find $f^{-1}(9) =$____

2. Originally Posted by viet
dont know how to start this:

$f(x) = 5 + 5 x + 4 e^x$

find $f^{-1}(9) =$____
There is a theorem that says:

Let $f$ be a one-to-one continuous function on an open interval $I$. If $f$ is differentiable at $x_0 \in I$ and if $f'(x_0) \neq 0$, then $f^{-1}$ is differentiable at $y_0 = f(x_0)$ and

$\left( f^{-1} \right)^{ \prime} (y_0) = \frac {1}{f'(x_0)}$

Note here, that $f(0) = 9$

Let $x_0 = 0$ and $y_0 = f(0) = 9$

Then $\left( f^{-1} \right)^{ \prime} (9) = \frac {1}{f'(0)}$

Now continue

3. Originally Posted by viet
dont know how to start this:

$f(x) = 5 + 5 x + 4 e^x$

find $f^{-1}(9) =$____
$f(x)$ is an increasing function so if there is a solution to:

$f(x)=9$

it is unique. Now $f(0)=9$, so $f^{-1}(9) = 0$

RonL

4. Originally Posted by CaptainBlack
$f(x)$ is an increasing function so if there is a solution to:

$f(x)=9$

it is unique. Now $f(0)=9$, so $f^{-1}(9) = 0$

RonL
well that was simple enough. anyway, my theorem is bound to come in handy at some point in the future