# Math Help - graphs -- urgent

1. ## graphs -- urgent

it'd be great if anyone could help me on this, im kinda stuck..

Okay, so there's a diagram, and it shows the graphs of f(x) = 1 + e^2x and g(x) = 10x + 2, where 0 [FONT='Times New Roman','serif']≤ x [FONT='Times New Roman','serif']≤ 1.5[/FONT][/FONT]
[FONT='Times New Roman','serif'][/FONT]
(a) (i) write down an expression for the vertical distance p between the graphs of f and g.
(ii) given that p has a maximum value for 0 [FONT='Times New Roman','serif']≤ x [FONT='Times New Roman','serif']≤ 1.5, find the value of x at which this occurs. [/FONT][/FONT]

(b) (i) i dont know if you guys could help me out on this one as well without the diagram, but here it is. so the graph of y = f(x) is there, and it says that when x = a and y = 5, what is f^-1(x).
(ii) hence show that a - ln2

thanks a lot!

2. The distance between the graphs is $p(x)=10x-e^{2x}+1$
$p'(x)=10-2e^{2x}$. Solving the equation $p'(x)=0$ we get $\displaystyle x=\frac{1}{2}\ln 5$ which is a point of maximum.
Then the maximum of $p$ is $p\left(\frac{1}{2}\ln 5\right)=5\ln 5-4$.

To find $f^{-1}$, from the equality $f(x)=y$ we must represent $x$ as a fucntion of $y$.
So $f(x)=y\Rightarrow 1+e^{2x}=y\Rightarrow e^{2x}=y-1\Rightarrow x=\frac{1}{2}\ln (y-1)$.
Then $\displaystyle f^{-1}(x)=\frac{1}{2}\ln (x-1)$

$f(a)=5\Leftrightarrow 1+e^{2a}=5\Leftrightarrow e^{2a}=4\Leftrightarrow 2a=2\ln 2\Leftrightarrow a=\ln 2$