The function f(x) = sin x (1 + cos x) is maxi | vedclass

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

Question

(C) Given $f(x) = \sin x (1 + \cos x) = \sin x + \sin x \cos x = \sin x + \frac{1}{2} \sin 2x$. Find the first derivative: $f'(x) = \cos x + \cos 2x$.

Vedclass · Accepted Answer

$x = \pi / 3$

Vedclass · Answer

(C) Given $f(x) = \sin x (1 + \cos x) = \sin x + \sin x \cos x = \sin x + \frac{1}{2} \sin 2x$. Find the first derivative: $f'(x) = \cos x + \cos 2x$. Set $f'(x) = 0$ for critical points: $\cos x + (2 \cos^2 x - 1) = 0$. Rearranging gives $2 \cos^2 x + \cos x - 1 = 0$. Factoring the quadratic: $(2 \cos x - 1)(\cos x + 1) = 0$. This gives $\cos x = 1/2$ or $\cos x = -1$. For $x \in [0, \pi]$,$\cos x = 1/2 \implies x = \pi / 3$ and $\cos x = -1 \implies x = \pi$. Find the second derivative: $f''(x) = -\sin x - 2 \sin 2x$. At $x = \pi / 3$: $f''(\pi / 3) = -\sin(\pi / 3) - 2 \sin(2\pi / 3) = -\frac{\sqrt{3}}{2} - 2(\frac{\sqrt{3}}{2}) = -\frac{3\sqrt{3}}{2} Since $f''(\pi / 3) < 0$,the function has a local maximum at $x = \pi / 3$.

The function $f(x) = \sin x (1 + \cos x)$ is maximum at $x = \dots$

Similar Questions

If $f(x)=\sin x+2 \cos ^{2} x$ for $\frac{\pi}{4} \leq x \leq \frac{3 \pi}{4}$, then $f$ attains its

$A$ circular sector of perimeter $60 \ m$ with maximum area is to be constructed. The radius of the circular arc in meters must be: (in $m$)

The local maximum value of the function $f(x) = \left(\frac{2}{x}\right)^{x^{2}}$,$x > 0$,is

If $m$ and $n$ respectively are the number of local maximum and local minimum points of the function $f(x) = \int_{0}^{x^{2}} \frac{t^{2}-5t+4}{2+e^{t}} dt$,then the ordered pair $(m, n)$ is equal to

The slant height of a right circular cone is $3 \text{ cm}$. The height of the cone for maximum volume is

Vedclass Test Series

Exam Paper Generator

Online Exam Module