In which of the following intervals does f(x) | vedclass

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

Question

(A) We want to find the interval where $f(x)$ increases less rapidly than $g(x)$,which means $f'(x) < g'(x)$.Let $h(x) = g(x) - f(x) = -2x^3 + 9x^2 -

Vedclass · Accepted Answer

$(1, 2)$

Vedclass · Answer

(A) We want to find the interval where $f(x)$ increases less rapidly than $g(x)$,which means $f'(x) Let $h(x) = g(x) - f(x) = -2x^3 + 9x^2 - 12x + 6$.We want $h'(x) > 0$ for $f(x)$ to increase less rapidly than $g(x)$.$h'(x) = -6x^2 + 18x - 12 = -6(x^2 - 3x + 2) = -6(x - 1)(x - 2)$.For $h'(x) > 0$,we need $(x - 1)(x - 2) This inequality holds when $x \in (1, 2)$.Thus,in the interval $(1, 2)$,$f(x)$ increases less rapidly than $g(x)$.

In which of the following intervals does $f(x) = 2x^3$ increase less rapidly than $g(x) = 9x^2 - 12x + 6$?

Similar Questions

Let $h(x) = f(x) - \{f(x)\}^2 + \{f(x)\}^3$ for every real number $x$,then

Let $g(x)=3 f\left(\frac{x}{3}\right)+f(3-x)$ and $f^{\prime \prime}(x)>0$ for all $x \in(0,3)$. If $g$ is decreasing in $(0, \alpha)$ and increasing in $(\alpha, 3)$,then $8 \alpha$ is

The value of $k$ such that the function $f(x) = \sin x - \cos x - kx + b$ is strictly decreasing for all real $x$ is

The interval in which $y=x^{2} e^{-x}$ is increasing is

Function $f(x) = |\sin x|$,$x \in \left(-\frac{\pi}{2}, 0\right)$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module