The function f(x) = (x - 2)|x - 3| is monoton | vedclass

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

Question

(A) The function is defined as $f(x) = \begin{cases} (x-2)(x-3) = x^2 - 5x + 6, & x \ge 3 \ (x-2)(-(x-3)) = -(x^2 - 5x + 6) = -x^2 + 5x - 6, & x < 3

Vedclass · Accepted Answer

$(-\infty, \frac{5}{2}) \cup (3, \infty)$

Vedclass · Answer

(A) The function is defined as $f(x) = \begin{cases} (x-2)(x-3) = x^2 - 5x + 6, & x \ge 3 \ (x-2)(-(x-3)) = -(x^2 - 5x + 6) = -x^2 + 5x - 6, & x For $x \ge 3$,$f'(x) = 2x - 5$. Since $x \ge 3$,$2x - 5 \ge 2(3) - 5 = 1 > 0$. Thus,$f(x)$ is increasing on $[3, \infty)$.For $x  0$,which means $-2x + 5 > 0$,or $x Combining these intervals,the function is monotonically increasing in $(-\infty, \frac{5}{2}) \cup (3, \infty)$.

The function $f(x) = (x - 2)|x - 3|$ is monotonically increasing in:

Similar Questions

Find the intervals in which the function $f$ given by $f(x)=2x^{3}-3x^{2}-36x+7$ is
$(a)$ increasing
$(b)$ decreasing

The function $f(x) = \tan x - x$ is:

If $f(x) = x^3 + bx^2 + cx + d$ and $0 < b^2 < c$,then in $(-\infty, \infty)$:

What type of function is $f(x) = \frac{e^{2x} - 1}{e^{2x} + 1}$?

Let $f(x)=x^{13}+x^{11}+x^{9}+x^{7}+x^{5}+x^{3}+x+19$. Then, $f(x)=0$ has

Vedclass Test Series

Exam Paper Generator

Online Exam Module