How to solve for x ?
Some trial and error might help here.Originally Posted by totalnewbie
Trial 1
Assume the $\displaystyle x=\sqrt{e}$ point is the tangency point on $\displaystyle f(x)$
Then we can find the slope of the tangent at $\displaystyle x=\sqrt{e}$, and
the corresponding value of $\displaystyle f(x)$, which will allow us to find
the equation of the tangent. Now check that this line is a tangent
to $\displaystyle g(x)$, if it is our job is done, if not proceed to Trial 2
Trial 2
Assume the $\displaystyle x=\sqrt{e}$ point is the tangency point on $\displaystyle g(x)$,
then proceed as in Trial 1, but with the roles of $\displaystyle f$ and $\displaystyle g$
interchanged.
RonL
Hello,Originally Posted by totalnewbie
if there exists a common tangent then the gradient of both functions must be equal. So first calculate the drivative of both functions:
$\displaystyle \frac{df}{dx}=\frac{1}{e} \cdot x$
$\displaystyle \frac{dg}{dx}=\frac{1}{x}$
Both are equal: $\displaystyle \frac{1}{e} \cdot x=\frac{1}{x}$
Solve for x and you'll get: $\displaystyle x=\sqrt{e}$
Complete the coordinates:
$\displaystyle f(\sqrt{e})=\frac{1}{2}$
$\displaystyle g(\sqrt{e})=\frac{1}{2}$
That means that the graphs of both functions have one common point: the tangent point. There exists one tangent (that's a special case, normally there must be two tangents!)
Use the point-slope-formula: $\displaystyle t: y=\frac{1}{\sqrt{e}} \cdot (x-\sqrt{e})+\frac{1}{2}$
I've attached a drawing to demonstrate my results.
Greetings
EB
Hello,Originally Posted by totalnewbie
you have typed the function f like this: x^2/2e. Then your computer will calculate $\displaystyle \frac{x^2}{2} \cdot e$.
You have to use parantheses: $\displaystyle \frac{x^2}{(2 \cdot e)}$
Good luck.
Greetings
EB