Question

For the graph of $f\left(x\right)=\ln \left(x^2–4x+5\right),$ which of the following is/are true:

• A. It has a horizontal asymptote.
• B. It has inflection poins at $(2,\ln 2)$ and $(3,\ln 2)$
• C. It has inflection poins at $(1,\ln 2)$ and $(3,\ln 2)$
• D. It has minimum at $x=2$

Answer 1

Answer: (C), (D)

It has minimum at $x=2$

Domain of the function is $x\in R$
Since $x^2–4x+5>0,\Rightarrow D=16–4\cdot 5=–4<0.$
$x-$ and $y-$ intercepts:
$f\left(x\right)=0,\Rightarrow \ln \left(x^2–4x+5\right)=0,$
$\Rightarrow x^2–4x+5=1,\Rightarrow x^2–4x+4=0,$
$\Rightarrow {\left(x–2\right)}^2=0,\Rightarrow x=2.$
To check for horizontal asymptote, we need to compute the following limit:
$\underset{x\to \pm \infty }{lim}f\left(x\right)=\underset{x\to \pm \infty }{lim}\ln \left(x^2–4x+5\right)=+\infty .$
Hence, the function has no horizontal asymptotes. Similarly, we can make sure that there are no oblique asymptotes.
Using L’Hopital’s rule, we have
$k=\underset{x\to \pm \infty }{lim}\frac{f\left(x\right)}{x}=\underset{x\to \pm \infty }{lim}\frac{\ln \left(x^2–4x+5\right)}{x}$
$=\left[\frac{\infty }{\infty }\right]$ form.
$=\underset{x\to \pm \infty }{lim}\frac{{\left(\ln \left(x^2–4x+5\right)\right)}^{\prime }}{x^{\prime }}$
$=\underset{x\to \pm \infty }{lim}\frac{\frac{2x–4}{x^2–4x+5}}{1}=\underset{x\to \pm \infty }{lim}\frac{2x–4}{x^2–4x+5}=0.$
The first derivative is given by
$f^{\prime }\left(x\right)={\left(\ln \left(x^2–4x+5\right)\right)}^{\prime }=\frac{2x–4}{x^2–4x+5}.$

$f^{\prime }\left(x\right)=0,\Rightarrow \frac{2x–4}{x^2–4x+5}=0,$
$\Rightarrow \{\begin{array}{c}2x–4=0\\ x^2–4x+5\ne 0\end{array},\Rightarrow x=2.$


As we can see from the sign chart, $x=2$ is a point of local minimum. Its $y-$value is
$f\left(2\right)=\ln \left(2^2–4\cdot 2+5\right)=\ln 1=0.$
Hence $f\left(x\right)\ge 0$ for all $x\in R$
Now,
$f^{\prime \prime }\left(x\right)={\left(\frac{2x–4}{x^2–4x+5}\right)}^{\prime }=\frac{–2x^2+8x–6}{{\left(x^2–4x+5\right)}^2}.$
$f^{\prime \prime }\left(x\right)=0,\Rightarrow \frac{–2x^2+8x–6}{{\left(x^2–4x+5\right)}^2}=0,\Rightarrow \{\begin{array}{c}–2x^2+8x–6=0\\ {\left(x^2–4x+5\right)}^2\ne 0\end{array},$
$\Rightarrow \left(x–1\right)\left(x–3\right)=0,$
$\Rightarrow x_1=1,x_2=3.$
It follows from the sign chart that both these points are points of inflection.
$f\left(1\right)=\ln \left(1^2–4\cdot 1+5\right)=\ln 2;$
$f\left(3\right)=\ln \left(3^2–4\cdot 3+5\right)=\ln 2.$
Thus, the function has the following inflection points: $\left(1,\ln 2\right)$ and $\left(3,\ln 2\right)$


Now, we can draw the graph of the function and it can be plotted as shown below:


Clearly, it has minimum at $x=2$ and it does not have any asymtote.