What is one of the maximum values of f(x) = x | vedclass

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

Question

(C) Let $f(x) = x + \sin(2x)$.First,find the derivative: $f'(x) = 1 + 2\cos(2x)$.Set $f'(x) = 0$ to find critical points: $1 + 2\cos(2x) = 0 \Rightarr

Vedclass · Accepted Answer

$\frac{\pi}{3} + \frac{\sqrt{3}}{2}$

Vedclass · Answer

(C) Let $f(x) = x + \sin(2x)$.First,find the derivative: $f'(x) = 1 + 2\cos(2x)$.Set $f'(x) = 0$ to find critical points: $1 + 2\cos(2x) = 0 \Rightarrow \cos(2x) = -\frac{1}{2}$.In the interval $[0, 2\pi]$,$2x$ can be $\frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}$.Thus,$x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$.Find the second derivative: $f''(x) = -4\sin(2x)$.Check the sign of $f''(x)$ at critical points:For $x = \frac{\pi}{3}$,$f''(\frac{\pi}{3}) = -4\sin(\frac{2\pi}{3}) = -4(\frac{\sqrt{3}}{2}) = -2\sqrt{3} For $x = \frac{2\pi}{3}$,$f''(\frac{2\pi}{3}) = -4\sin(\frac{4\pi}{3}) = -4(-\frac{\sqrt{3}}{2}) = 2\sqrt{3} > 0$ (Local Minimum).Calculating the value at $x = \frac{\pi}{3}$: $f(\frac{\pi}{3}) = \frac{\pi}{3} + \sin(\frac{2\pi}{3}) = \frac{\pi}{3} + \frac{\sqrt{3}}{2}$.

What is one of the maximum values of $f(x) = x + \sin(2x)$ in the interval $[0, 2\pi]$?

Similar Questions

The minimum value of $ax + by$ where $xy = c^2$ is

For all $x \in \mathbb{R}$, the minimum value $\frac{1}{3}$ and the maximum value $3$ of $f(x) = \frac{x^2+x+1}{x^2-x+1}$ occur at $l$ and $m$ respectively. Then $l+m$ is equal to:

Let $f(x) = \int_{0}^{x} e^{x+t} dt$. Then the abscissa of the point where the tangent to $f(x)$ is parallel to the $x$-axis is:

The function $f(x) = 2x^3 - 15x^2 + 36x + 4$ is maximum at $x=$ ......

Let $f(x) = (x^2 - 1)^n (x^2 + x + 1)$. Then $f(x)$ has a local extremum at $x = 1$ when:

Vedclass Test Series

Exam Paper Generator

Online Exam Module