Show that a function has exactly one real root I have a function : $$f(x) = e^{3x}+4e^{2x}-3e^x=0$$ I have to show that there is exactly one real root.
My effort
$f$ is continuous on $\mathbb{R}$
$f(-1) = e^{-3}+4e^{-2}-3e^{-1} \approx =-0.512$
$f(0) = e^{0}+4e^{0}-3e^{0}=2$
So by the Intermediate Value Theorem, we know that there must be a number $c \in [−1, 0]$, where $f(c) = 0.$
So I call $c$ my solution to the original equation and suppose there exists another one; I call this one $d$. Then either $d > c$ or $d < c$. Then $f(x)$ is a continuous function on $[c,d]$ and differentiable on $(c,d)$ (it is differentiable everywhere, actually). By Rolle’s Theorem, there then must exist a point $r \in (c,d)$, where $f′(r) = 0$.
Doing so $$f'(x) = \mathrm{e}^x\cdot\left(3\mathrm{e}^{2x}+8\mathrm{e}^x-3\right)$$
substituting $u = e^x$ I take : $$ u(3u^2+8u-3)$$ that have two roots $u=1/3$ and $u=-3$ substituting back now I take $x= \log(1/3)$ and $x=-\log(3)$.I have two roots not one.
Am I correct ?
 A: For a differentiable function $f$, the existence of a point $r$ where $f'(r)=0$ is a necessary condition for the function to have two roots (points $x$ where $f(x)=0$).
This even if the function has at least one root. As an easy example, take $f(x)=x^3-3x+3$. This has a root by the intermediate value theorem, and two points where the derivative vanishes, namely $-1$ and $1$. Since $f(-1)=5$ and $f(1)=1$, we can conclude that the function has only one root.
Your calculations are not wrong, but you make the wrong deduction from the gathered data.
You have $f'(x)=e^x(3e^{2x}+8e^x-3)$ and the derivative vanishes where
$e^x=1/3$, so for $x=-\log3$. You can check that this is a point of minimum and
$$
f(-\log3)=\frac{1}{27}+\frac{4}{9}-1=-\frac{14}{27}<0
$$
We also have

*

*$f(0)>0$,

*$\lim\limits_{x\to-\infty}f(x)=0$,

*$\lim\limits_{x\to\infty}f(x)=\infty$.

Thus a second root requires two points where the derivative vanishes.

Of course this can be avoided by observing that
$$
f(x)=e^x(e^{2x}+4e^x-3)=e^x(e^x-2+\sqrt{7})(e^x-2-\sqrt{7})
$$
and the first two factors are positive. But I guess that the purpose of the exercise was to use calculus and not algebra.
A: As suggested by MartinR in the comments, there is actually no need to use real analysis for this question. If we substitute $t=e^x>0$, we have to show that the equation $$t^3+4t^2-3t=0$$ has exactly one positive root. But this easily factors as $t(t^2+4t-3)=0$, so we are left with the equation $t^2+4t-3=0$, whose roots are $$t_{1, 2}=-2\pm \sqrt{7}.$$
Since $t_1=-2-\sqrt{7}<0$, only $t_2=\sqrt{7}-2$ works for us and the desired conclusion follows (note that we can actually find explicitly the unique real root of your equation).
A: The question is also equivalent to asking for the number of (real) intersections of the curve for $ \ g(x) \ = \ e^{3x} + 4e^{2x} \ = \ e^{2x}·(e^x + 4 ) \ $ with the curve for $ \  h(x) \ = \  3e^x \ \ . \ $  We observe that $ \ g(0) \ = \ 1 + 4 \ > \ h(0) \ = \ 3 \ \ $ and that $ \ h(-1) \ = \ 3·e^{-1} \ =  \ \frac{3e}{e^2}  \ > \ \frac{8}{e^2} \ > \ \frac{e^{-1} \  + \ 4}{e^2} \ = \ g(-1) \ \ . \  $  Both $ \ g(x) \ $ and $ \ h(x) \ $ are continuous "everywhere", so the Intermediate Value Theorem guarantees that there is a value $ \ -1 \ < \ c \ < \ 0 \ $ for which $ \ g(c) \ = \ h(c) \ : \ $ there is certainly at least one real intersection of the curves.
We can show readily enough that for $ \ x \ \ge \ 0 \ \ ,  \ e^x \ > \ 3 \ \Rightarrow \ g(x) \ > \ e^{2x} \ = \ e^x·e^x  \ > \ 3e^{ x} \ = \ h(x) \ \ . \  $ For $ \ x \ \le \ -1 \ \ , \ $ we can make an argument similar to the one in the first paragraph to show that $ \ h(-|x|) \ = \ 3·e^{-|x|}   \ > \ \frac{e^{-|x|} \  + \ 4}{e^{2|x|}} \ = \ g(-|x|) \ \ . \  $
As for the first derivatives $ \ g'(x) \ = \ e^{2x}·(3e^x + 8) \ $ and $ \ h'(x) \ = \ 3e^x \ = \  h(x) \ \ , \ $ we find that
$  g'(-|x|) \ = \ \frac{3e^{-|x|} \  + \ 8}{e^{2·|x|}} \ > \  3·e^{-|x|} \ = \ h'(-|x|) \ \ $ for $ \ x \ < \ 0 \ \ . \ $  So $ \ g(x) \ $ "grows faster" than $ \ h(x) \ \   $ everywhere, telling us that the    intersection at $ \ x \ = \ c \ $ is the only one there is for these curves (and why you don't obtain a second zero of $ \ f'(x) \ $ in order to apply Rolle's Theorem).  [ Alexdanut and egreg have shown the location of the sole intersection to be $ \ c \ = \ \ln \ (\sqrt7 - 2) \ \approx \ -0.437 \ \ . \ ] $
