1
$\begingroup$

So I'm trying to solve a system of equations and I checked some other guys solution and he divides the function by the derivate, like so: $f(x)/f'(x)$.

Find the values of the real constant $k$ for which this holds true:
$$ \log(2x) \leq kx \leq e^{x/2}, \quad x > 0 $$ I convert this into 2 systems of equations: $$ f(x)=\log(2x)=kx \ and \ f'(x)=1/x=k \\ f(x)=e^{x/2}=kx \ and \ f'(x)=\frac{e^{x/2}}{2}=k $$ Then I got stuck,checked some other guys solution and he did this: $f(x)/f'(x)$ for both equations. Why?

Sorry for the long winded explanation but I really can't figure out why it's $f(x)/f'(x)$.

Progress

Figured it out. Dividing $f(x)$ with $f'(x)$ is just another way of creating an expression for x and inserting it into the other equation in the system.

$\endgroup$
6
  • $\begingroup$ I can't either. Nobody guarantees that this other guy's solution is correct. Did you solve this problem? $\endgroup$ Nov 15, 2014 at 20:39
  • $\begingroup$ @TZakrevskiy His solution is correct, I checked with Mathematica $\endgroup$
    – Lightvvind
    Nov 15, 2014 at 21:09
  • $\begingroup$ Well, unless you provide us with a detailed solution involving $f(x)/f'(x)$, I'm inclined to believe that that guy obtained a correct answer by an incorrect method. $\endgroup$ Nov 15, 2014 at 21:20
  • $\begingroup$ $f´(x)$ is equal to k in both cases, insert that into $f(x)$ and solve for x. Then solve for k. $\endgroup$
    – Lightvvind
    Nov 15, 2014 at 21:23
  • $\begingroup$ And where does the fraction $\frac{f(x)}{f'(x)}$ come into play in this method? $\endgroup$ Nov 15, 2014 at 21:29

1 Answer 1

1
$\begingroup$

You are given two functions, $f(x) = e^{x/2}$ and $g(x) = \log(2x)$ for $x > 0$. The problem is to find out which lines through the origin $h(x) = kx$ pass in between these curves for all values of $x > 0$. The first thing to do, is to sketch a plot of $f$ and $g$. This plot shows that there is indeed a gap through which straight lines through the origin will pass.

The line with the maximum allowed value of k will pass just underneath $f$. Therefore we must find the point $(x_0, f(x_0))$ on $f$ where the tangent passes through the origin. Now the derivative of $f(x)$ is given by $f'(x) = (1/2)e^{x/2}$. Therefore the tangent for $x=x_0$ is given by the equation:

tangent to $f = (x_0, e^{x_0/2}) + p (1, (1/2)e^{x_0/2}) $

The intersection with the $y$-axis is obtained by setting $p = -x_0$. The corresponding $y$-coordinate becomes $(1 - x_0/2) e^{x_0/2}$. Therefore $x_0 = 2$ is the solution. Substituting this value in the tangent yields $k = e/2$. This is the upper bound for $k$.

For $g(x)$ we find the derivative $g'(x) = 2/x.$ The tangent through a point $x0$ on $g$:

tangent to $g = (x_0, \log(2x_0)) + q (1, 2/x_0)$

The intersection with the y-axis occurs when $q = -x_0$. The corresponding $y$-value is $\log(2x_0) - 2$. Solving yields $x_0 = (1/2)e^2$. Substitution in the tangent yields $k = 4/e^2$. This the lower bound for $k$.

In summary, we find that the set of straight lines through the origin that pass in between $f$ and $g$ are given by $y = kx$ with $k$ in the interval $(4/e^2, e/2)$.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .