Find the values of x for which y=[x(x-2)]^{2} | vedclass

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

Question

(A) We have,$y = [x(x-2)]^2 = (x^2 - 2x)^2$. To find the intervals where the function is increasing,we calculate the derivative: $\frac{dy}{dx} = 2(x^

Vedclass · Accepted Answer

$x \in (0, 1) \cup (2, \infty)$

Vedclass · Answer

(A) We have,$y = [x(x-2)]^2 = (x^2 - 2x)^2$. To find the intervals where the function is increasing,we calculate the derivative: $\frac{dy}{dx} = 2(x^2 - 2x)(2x - 2) = 4x(x-2)(x-1)$. Setting $\frac{dy}{dx} = 0$,we get the critical points: $x = 0, 1, 2$. These points divide the real line into intervals $(-\infty, 0), (0, 1), (1, 2), (2, \infty)$. Testing the sign of $\frac{dy}{dx}$ in each interval: $1$. For $x \in (-\infty, 0)$,$\frac{dy}{dx} $2$. For $x \in (0, 1)$,$\frac{dy}{dx} > 0$ (increasing). $3$. For $x \in (1, 2)$,$\frac{dy}{dx} $4$. For $x \in (2, \infty)$,$\frac{dy}{dx} > 0$ (increasing). Thus,the function is increasing for $x \in (0, 1) \cup (2, \infty)$.

Find the values of $x$ for which $y=[x(x-2)]^{2}$ is an increasing function.

Similar Questions

Where does $f(x) = x + \sqrt{1 - x}, 0 < x < 1$ decrease?

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

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

If $g(x) = \frac{1}{6} f(3 x^2 - 1) + \frac{1}{2} f(1 - x^2), \forall x \in R$,where $f''(x) > 0, \forall x \in R$. Then $g(x)$ is increasing in the interval

Statement $-1:$ The function $f(x) = x^2(e^x + e^{-x})$ is increasing for all $x > 0.$
Statement $-2:$ The functions $g(x) = x^2e^x$ and $h(x) = x^2e^{-x}$ are increasing for all $x > 0$ and the sum of two increasing functions in any interval $(a, b)$ is an increasing function in $(a, b).$

Vedclass Test Series

Exam Paper Generator

Online Exam Module