Question

Let $f:R\to R$ be given by
$f\left(x\right)=\{\begin{array}{c}{x}^5+5x^4+10x^3+10x^2+3x+1,x<0;\\ x^2-x+1,0\le x<1;\\ \frac{2}{3}{x}^3-4x^2+7x-\frac{8}{3},1\le x<3;\\ (x-2){\log }_e(x-2)-x+\frac{10}{3},x\ge 3.\end{array}$
Then which of the following option is/are correct?

• A. $f$ is increasing on $(-\infty ,0)$
• B. $f^{\prime }$ has a local maximum at $x=1$
• C. $f$ is onto
• D. $f^{\prime }$ is NOT differentiable at $x=1$

Answer 1

Answer: (B), (C), (D)

$f^{\prime }$ is NOT differentiable at $x=1$

$f\left(x\right)=\{\begin{array}{c}(x+1{)}^5-2x,x<0;\\ x^2-x+1,0\le x<1;\\ \frac{2}{3}{x}^3-4x^2+7x-\frac{8}{3},1\le x<3;\\ (x-2){\log }_e(x-2)-x+\frac{10}{3},x\ge 3.\end{array}$

$f^{\prime }\left(x\right)=\{\begin{array}{c}5(x+1{)}^4-2,x<0;\\ 2x-1,0\le x<1;\\ 2x^2-8x+7,1\le x<3;\\ {\log }_e(x-2),x\ge 3.\end{array}$

$f^{\prime \prime }\left(x\right)=\{\begin{array}{c}20(x+1{)}^3,x<0;\\ 2,0\le x<1;\\ 4x-8,1\le x<3;\\ \frac{1}{x-2},x\ge 3;\end{array}$

For $x=1$
$f^{\prime \prime }\left(1^{+}\right)=\underset{x\to 1^{+}}{lim}4(1+h)-8\Rightarrow \underset{h\to 0}{lim}4h-4=-4$

$f^{\prime \prime }\left(1^{-}\right)=\underset{x\to 1^{-}}{lim}2=2$
As $f^{\prime \prime }\left(1^{+}\right)\ne f^{\prime \prime }\left(1^{-}\right)$
$f^{\prime }\left(x\right)$ is not differntiable at $x=1$
And from the graph of $f^{\prime }\left(x\right)$
$\therefore f^{\prime }\left(x\right)$ has local maximum at $x=1$.
For $x<0,f\left(x\right)$ is polynomial function so $f\left(x\right)$ is continuous in $(-\infty ,0)$
Also, $\underset{x\to -\infty }{lim}f\left(x\right)=-\infty$ and $\underset{x\to 0^{-}}{lim}f\left(x\right)=1$
Hence, $(-\infty ,1)\subset Rangeoff\left(x\right)\text{in}(-\infty ,0)$
For $x\ge 3,f\left(x\right)$ is continuous so domain of $f\left(x\right)\text{is}R.$

$\underset{x\to \infty }{lim}f\left(x\right)=\infty$ and $\underset{x\to 3}{lim}f\left(x\right)=\frac{1}{3}$
$[\frac{1}{3},\infty )\subset Rangeoff\left(x\right)in[3,\infty )$
Hence, range of all the function are subset of range $f\left(x\right).$

Therefore range and domain of $f\left(x\right)$ is $R(-\infty ,\infty )$ therefore $f$ is onto function.

For maxima and minima,
$f^{\prime }\left(x\right)=0\Rightarrow 5(x+1{)}^4-2=0$
$x=-1\pm \sqrt[4]{4\frac{2}{5}}$
so, $f^{\prime }\left(x\right)=0$ changes sign in $(-\infty ,0)$
hence $f\left(x\right)$ is not increaing on $(-\infty ,0)$