The function f(x) = x + sin x has | vedclass

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

Question

(C) Given the function $f(x) = x + \sin x$. First,we find the derivative: $f'(x) = 1 + \cos x$. Setting $f'(x) = 0$,we get $1 + \cos x = 0$,which impl

Vedclass · Accepted Answer

Neither maximum nor minimum

Vedclass · Answer

(C) Given the function $f(x) = x + \sin x$. First,we find the derivative: $f'(x) = 1 + \cos x$. Setting $f'(x) = 0$,we get $1 + \cos x = 0$,which implies $\cos x = -1$. This occurs at $x = (2n+1)\pi$ for any integer $n$. Now,we check the second derivative: $f''(x) = -\sin x$. At $x = \pi$,$f''(\pi) = -\sin(\pi) = 0$. Since the second derivative is zero,we check the third derivative: $f'''(x) = -\cos x$. At $x = \pi$,$f'''(\pi) = -\cos(\pi) = -(-1) = 1 
eq 0$. Since the first non-zero derivative at the critical point is of an odd order (the third derivative),the function $f(x)$ has a point of inflection at $x = \pi$ and does not have a local maximum or local minimum.

The function $f(x) = x + \sin x$ has

Similar Questions

The function $f(x) = \frac{x^2 - 2}{\sqrt{1 + x^2}}$:

The abscissae of the points,where the tangent to the curve $y = x^3 - 3x^2 - 9x + 5$ is parallel to the $x$-axis,are

Find the absolute maximum value and the absolute minimum value of the function given by $f(x) = \sin x + \cos x$ for $x \in [0, \pi]$.

$A$ wire $40 \text{ m}$ in length is to be cut into two pieces. One piece is formed into a square and the other piece into a circle. What should be the lengths of the two pieces so that the combined area of the square and the circle is minimum?

Vedclass Test Series

Exam Paper Generator

Online Exam Module