For what interval is the function f(x) = sin | vedclass

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

Question

(C) Given $f(x) = \sin x - \cos x$.To find the interval where the function is strictly increasing,we find the derivative $f'(x)$.$f'(x) = \cos x + \si

Vedclass · Accepted Answer

$x \in (-\pi /4, 3\pi /4)$

Vedclass · Answer

(C) Given $f(x) = \sin x - \cos x$.To find the interval where the function is strictly increasing,we find the derivative $f'(x)$.$f'(x) = \cos x + \sin x$.We can rewrite this as $f'(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x ight) = \sqrt{2} \sin(x + \pi /4)$.The function is strictly increasing when $f'(x) > 0$.$\sqrt{2} \sin(x + \pi /4) > 0 \implies \sin(x + \pi /4) > 0$.Since $\sin 	heta > 0$ for $	heta \in (0, \pi)$,we have $0 Subtracting $\pi /4$ from all parts,we get $-\pi /4 Thus,the function is strictly increasing for $x \in (-\pi /4, 3\pi /4)$.

For what interval is the function $f(x) = \sin x - \cos x$ strictly increasing?

Similar Questions

If the interval in which the real-valued function $f(x) = \log \left(\frac{1+x}{1-x}\right) - 2x - \frac{x^3}{1-x^2}$ is decreasing is $(a, b)$, where $|b-a|$ is maximum, then $\frac{a}{b} =$

The equation $x^3+x-1=0$ has

For what values of $K$ is the function $f(x) = Kx^3 + 9x^2 + 9x + 3$ an increasing function on $R$?

If $f(x) = \frac{\sin x + b \cos x}{\sin x + 4 \cos x}$ is a strictly increasing function,then:

What type of function is $f(x) = \frac{e^{2x} - 1}{e^{2x} + 1}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module