1. ## inverse

dont know how to start this:

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

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

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

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

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

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

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

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

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

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

Now continue

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

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

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

$\displaystyle f(x)=9$

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

RonL

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

$\displaystyle f(x)=9$

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

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