Question

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

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

Answer 1

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

$f(x)=\{\begin{array}{ll}{x}^5+5x^4+10x^3+10x^2+3x+1,& x<0\\ x^2-x+1,& 0\le x<1;\\ \frac{2}{3}{x}^3-4x^2+7x-\frac{8}{3},& 1\le x<3;\\ (x-2)lo{g}_e(x-2)-x+\frac{10}{3},& x\ge 3.\end{array}$
Clearly f(x) is continuous at x = 0, 1 and 3
$f(x)=\{\begin{array}{ll}5x^4+20x^3+30x^2+20x+3,& \text{x}<0\\ 2x-1,& \text{0}<x<1\\ 2x^2-8x+7,& \text{1}<x<3\\ lo{g}_e(x-2),& \text{x}>3.\end{array}$
at $x=1,f^{\prime \prime }(1^{-})>0$ and $f^{\prime \prime }(1^{+})<0$
$\therefore f^{\prime }(x)$ has local maxima at $x=1$.
Option (A) is correct
and $f^{\prime \prime }\left(1^{-}\right)\ne f^{\prime \prime }\left(1^{+}\right)$
$\Rightarrow f^{\prime }$ is not differentiable at $x=1$
Option (B) is correct
$f(x)$ has range $\left(-\infty ,\infty \right)$.
$\therefore f$ is onto $\Rightarrow$ Option (C) is correct
For $x<0,f^{\prime }(x)=5x^4+20x^3+30x^2+20x+3.$
Here $f^{\prime }(-1)<0$
$\therefore f(x)$ is not monotonically increasing on $\left(-\infty ,0\right)$