1
$\begingroup$

I've been given this function:

$f(x) = -4\ln(e^x -2) + e^x + 4x$

I've been asked to find the intersection with the axis. I know the domain is $x \gt \ln(2)$ which means there's no intersection with the y axis.

Since I can't really calculate the value of x in: $-4\ln(e^x -2) + e^x + 4x = 0$. I'm assuming towards the fact that there's no intersection with the axis and I need to prove it which I'm having trouble with.

The farthest I got is that I derived the function twice and found out that $f''(x) \gt 0$ for all x in my domain. I know this means that $f'(x)$ is increasing. What else can I do with this? Am I even on the right track?

Thanks

  • 0
    If $f'$ is increasing and $f'$ is positive, then what is your function doing? – 2017-02-07
  • 0
    The thing is, I'm not sure f' is always positive (It might be and I haven't discovered it) – 2017-02-07
  • 0
    $f$ has its **global** minimum at $x_\min=\ln 4$ with $f(x_\min)=4 + \ln 16$! – 2017-02-07
  • 0
    @RonH But surely you can play around with it a bit... – 2017-02-07

3 Answers 3

0

If I consider $f(x)$ to be a real function, then it is defined only for $e^x > 2$. Notice that for large positive values of $x$, $\ln(e^x - 2) \sim x$. Thus, for large positive $x$, $f(x)$ is also positive. Hence it suffices to prove two things:

$(1)$ $$ \lim_{x\to 2^+} f(x) > 0 $$

$(2)$ The minima of $f(x)$ lies above the $x$-axis.

0

Let $v=e^x$. Then with just a bit of algebra, your equation becomes $$ \ln\left(1-\frac2{v}\right) = v/4 $$ IN the domain in question, the left side is always negative and the right side always positive; therefore the equation has no solution.

0

For $\log 2 < x \leq \log 3$ we have $0

For $x > \log 3,$ $\log (e^x - 2) < \log (e^x ) = x, $ so $$ 4 (x - \log (e^x - 2)) > 0 $$