A function is matched below against an interv | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) To determine if a function $f(x)$ is increasing on an interval,we check if $f'(x) \ge 0$ for all $x$ in that interval.$(a)$ For $f(x) = 3x^2 - 2x

Vedclass · Accepted Answer

$\left( -\infty, \frac{1}{3} \right]$ | $3x^2 - 2x + 1$

Vedclass · Answer

(A) To determine if a function $f(x)$ is increasing on an interval,we check if $f'(x) \ge 0$ for all $x$ in that interval.$(a)$ For $f(x) = 3x^2 - 2x + 1$,$f'(x) = 6x - 2$. Setting $f'(x) \ge 0$ gives $6x \ge 2$,so $x \ge \frac{1}{3}$. The function is increasing on $[\frac{1}{3}, \infty)$,not $(-\infty, \frac{1}{3}]$. Thus,option $(a)$ is incorrectly matched.$(b)$ For $f(x) = x^3 + 6x^2 + 6$,$f'(x) = 3x^2 + 12x = 3x(x + 4)$. $f'(x) \ge 0$ when $x \in (-\infty, -4] \cup [0, \infty)$. Thus,it is increasing on $(-\infty, -4]$.$(c)$ For $f(x) = x^3 - 3x^2 + 3x + 3$,$f'(x) = 3x^2 - 6x + 3 = 3(x - 1)^2$. Since $3(x - 1)^2 \ge 0$ for all $x \in \mathbb{R}$,the function is increasing on $(-\infty, \infty)$.$(d)$ For $f(x) = 2x^3 - 3x^2 - 12x + 6$,$f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x - 2)(x + 1)$. $f'(x) \ge 0$ when $x \in (-\infty, -1] \cup [2, \infty)$. Thus,it is increasing on $[2, \infty)$.

$A$ function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched?
Interval | Function

Similar Questions

The interval in which the function $f(x) = 2x^2 - \log x$,for $x > 0$ decreases is

The value of $K$ such that $f(x) = \sin x - \cos x - Kx + 5$ decreases for all positive real values of $x$ is given by

The function,$f(x)=(3x-7)x^{2/3}, x \in R,$ is increasing for all $x$ lying in

Let $f(x) = \cos \left(\frac{\pi}{x}\right), x \neq 0$. Assuming $k$ is an integer, which of the following is true?

For all real values of $x$,the increasing function $f(x)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$A$ function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched?Interval | Function