Let f(x) = e^x - e^{-x} + cos x,then f(x) is | vedclass

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

Question

(A) Given $f(x) = e^x - e^{-x} + \cos x$. First,we find the derivative $f'(x)$: $f'(x) = \frac{d}{dx}(e^x - e^{-x} + \cos x) = e^x + e^{-x} - \sin x$.

Vedclass · Accepted Answer

always increasing

Vedclass · Answer

(A) Given $f(x) = e^x - e^{-x} + \cos x$. First,we find the derivative $f'(x)$: $f'(x) = \frac{d}{dx}(e^x - e^{-x} + \cos x) = e^x + e^{-x} - \sin x$. We know that $e^x + e^{-x} = 2 \cosh x$. Since $e^x + e^{-x} \ge 2$ for all $x \in \mathbb{R}$ (by $AM$-$GM$ inequality) and $\sin x \le 1$ for all $x \in \mathbb{R}$,we have: $f'(x) = (e^x + e^{-x}) - \sin x \ge 2 - 1 = 1$. Since $f'(x) > 0$ for all $x \in \mathbb{R}$,the function $f(x)$ is always increasing.

Let $f(x) = e^x - e^{-x} + \cos x$,then $f(x)$ is

Similar Questions

Find the intervals in which the function $f(x) = -2x^{3} - 9x^{2} - 12x + 1$ is strictly increasing or strictly decreasing.

Find the intervals in which the function $f(x) = x^{2} + 2x - 5$ is strictly increasing or strictly decreasing.

Find the intervals in which the function given by $f(x) = \sin 3x, x \in \left[0, \frac{\pi}{2}\right]$ is:
$(a)$ increasing
$(b)$ decreasing.

If $f(x) = \frac{\lambda \sin x + 6 \cos x}{2 \sin x + 3 \cos x}$ is a strictly increasing function,then .......

Let the function $f(x) = \frac{x}{3} + \frac{3}{x} + 3$,$x \neq 0$ be strictly increasing in $(-\infty, \alpha_1) \cup (\alpha_2, \infty)$ and strictly decreasing in $(\alpha_3, \alpha_4) \cup (\alpha_4, \alpha_5)$. Then $\sum_{i=1}^5 \alpha_i^2$ is equal to :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module