Question

Let $f\,\mathbb{R} \to\mathbb{R}$ be given by
$f(x) = \begin{cases} x^5+5x^4+10x^3+10x^2+3x+1, & \quad \text{} x < 0 \\ x^2-x+1, & \quad \text{ } 0 \le x < 1; \\ \frac{2}{3}x^{3}-4x^{2}+7x-\frac{8}{3}, & \quad \text{} 1 \le x< 3;\\ (x - 2) log_e(x - 2) - x +\frac{10}{3}, & \quad \text{ } x \ge 3. \end{cases} $
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{cases} x^5+5x^4+10x^3+10x^2+3x+1, & \quad \text{} x < 0 \\ x^2-x+1, & \quad \text{ } 0 \le x < 1; \\ \frac{2}{3}x^{3}-4x^{2}+7x-\frac{8}{3}, & \quad \text{} 1 \le x< 3;\\ (x - 2) log_e(x - 2) - x +\frac{10}{3}, & \quad \text{ } x \ge 3. \end{cases} $
Clearly f(x) is continuous at x = 0, 1 and 3
$f(x) = \begin{cases} 5x^4 +20x^3 + 30x^2 + 20x + 3, & \quad \text x < 0 \\ 2x-1, & \quad \text 0 < x < 1 \\ 2x^2 - 8x+7, & \quad \text 1 < x < 3 \\ log_e(x-2), & \quad \text x > 3. \end{cases} $
at $x = 1, f'' (1^-) > 0$ and $f'' (1^+) < 0$
$\therefore f'(x)$ has local maxima at $x=1$.
Option (A) is correct
and $f ''\left(1^{-}\right)\ne f '' \left(1^+\right)$
$\Rightarrow f '$ 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 $x < 0, f'(x) = 5x^4 + 20x^3 + 30x^2 + 20x + 3.$
Here $f'(-1) < 0$
$\therefore f(x)$ is not monotonically increasing on $\left(-\infty, 0\right)$