Find and classify the critical points of $f(x) = \ln(e^{3x} −2e^{2x} + e^x + 2)$, then sketch its graph

Question

Question : Find and classify the critical points of $$f(x) = \ln(e^{3x} −2e^{2x} + e^x + 2),$$ then sketch its graph

I have found out that:

first derivative = $( 3e^{3x} - 4e^{2x} + e^x )(e^{3x} −2e^{2x} + e^x + 2)^{-1}$

Maximized at $x= -\ln3$ ; Minimized at $x=0$

Y-intercept $= \ln2$ ; No X-intercept

Second derivative = $(2e^x)(8e^x -8e^{2x} -2e^{3x} + e^{4x} -1) (e^{3x} −2e^{2x} + e^x + 2)^{-2}$

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The second derivative seems so complicated such that it's not easy to solve F''(x) =0 in order to find out the concavity of the function. Then I cannot sketch the graph of the function.

What should I do?

Answer 1

You do not have to use the second derivative test.

Using the fact that

$$ f^\prime(x)=\frac{(3e^x-1)(e^x-1)}{e^{3x} −2e^{2x} + e^x + 2} $$

with critical point $x=-\ln3$ and $x=0$ you can check the sign of $f^\prime(x)$ on each of the three intervals $(-\infty,-\ln3),\,(-\ln3,0),\,(0,\infty)$ by checking the values of $f^\prime(-\ln4),\,f^\prime(-\ln2),\,f^\prime(\ln2)$.

ADDENDUM: When sketching the graph, note that

1. $\displaystyle\lim_{x\to-\infty}\ln(e^{3x} −2e^{2x} + e^x + 2)=\ln(2)$
2. $\displaystyle\lim_{x\to\infty}\ln(e^{3x} −2e^{2x} + e^x + 2)=\displaystyle\lim_{x\to\infty}\ln(e^{3x})\sim3x$ which are, respectively, a horizontal and a slant asymptote.