Find the local maximum and local minimum valu | vedclass

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

Question

(A) Given function: $h(x) = \sin x + \cos x$ for $x \in (0, \frac{\pi}{2})$.First,find the derivative: $h'(x) = \cos x - \sin x$.Set $h'(x) = 0$ to fi

Vedclass · Accepted Answer

Local maximum value is $\sqrt{2}$ and no local minimum value exists.

Vedclass · Answer

(A) Given function: $h(x) = \sin x + \cos x$ for $x \in (0, \frac{\pi}{2})$.First,find the derivative: $h'(x) = \cos x - \sin x$.Set $h'(x) = 0$ to find critical points:$\cos x - \sin x = 0 \Rightarrow \sin x = \cos x \Rightarrow 	an x = 1 \Rightarrow x = \frac{\pi}{4}$.Find the second derivative: $h''(x) = -\sin x - \cos x$.Evaluate $h''(x)$ at $x = \frac{\pi}{4}$:$h''(\frac{\pi}{4}) = -\sin(\frac{\pi}{4}) - \cos(\frac{\pi}{4}) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$.Since $h''(\frac{\pi}{4}) The local maximum value is $h(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.Since the function is strictly concave down on the interval $(0, \frac{\pi}{2})$,there is no local minimum value in the given open interval.

Find the local maximum and local minimum values for the function $h(x) = \sin x + \cos x$ in the interval $0 < x < \frac{\pi}{2}$.

Similar Questions

Prove that the function $h(x) = x^{3} + x^{2} + x + 1$ does not have any local maxima or local minima.

The minimum value of ${x^2} + \frac{1}{1 + {x^2}}$ is at $x=$ ..........

In the interval $[0, 2\pi]$,the slope of the tangent to the function $f(x) = e^x \sin x$ is maximum at $x = \dots$

Find the semi-vertical angle of a right circular cone of a given slant height,if the volume of the cone is maximum.

Find the maximum value of $f(x) = x^3 - 12x^2 + 45x$ in the interval $[0, 7]$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module