Find the absolute extrema of the given function #f(x)= sinx+cosx# on interval #[0,2pi]#?

Answer 1

Isaiah Allen

#f_max=sqrt2#
#f_min=-sqrt2#

#f'(x)=cosx-sinx#

#f'(x)=0 <=> cosx-sinx=0#

#cosx/cosx-sinx/cosx=0 ^^ cosx !=0#

#tanx=1 ^^ cosx !=0#

#x=pi/4+kpi ^^ x != pi/2+mpi#

#x=pi/4+kpi#

On #[0,2pi]# :

#x=pi/4 vv x=(5pi)/4 #

#f_max=f(pi/4) = sin(pi/4)+cos(pi/4) = sqrt2/2+sqrt2/2=sqrt2#

#f_min=f((5pi)/4) = sin((5pi)/4)+cos((5pi)/4) = -sqrt2/2-sqrt2/2=-sqrt2#

#f_max=sqrt2# #f_min=-sqrt2#

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Answer 2

Emilia Aguilar

To find the absolute extrema of $f(x) = \sin(x) + \cos(x)$ on the interval $[0, 2\pi]$ , we first evaluate the function at the critical points and endpoints within the interval.

Evaluate $f(x)$ at the critical points: $f'(x) = \cos(x) - \sin(x)$ Set $f'(x) = 0$ to find critical points: $\cos(x) - \sin(x) = 0$ Solving for $x$ within the interval $[0, 2\pi]$ : $x = \frac{\pi}{4}, \frac{5\pi}{4}$
Evaluate $f(x)$ at the endpoints of the interval: $f(0) = \sin(0) + \cos(0) = 1$ $f(2\pi) = \sin(2\pi) + \cos(2\pi) = 1$
Evaluate $f(x)$ at the critical points: $f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \sqrt{2}$ $f\left(\frac{5\pi}{4}\right) = \sin\left(\frac{5\pi}{4}\right) + \cos\left(\frac{5\pi}{4}\right) = -\sqrt{2}$
Compare the values obtained in steps 2 and 3 to find the absolute extrema:
- Absolute maximum: $\sqrt{2}$ at $x = \frac{\pi}{4}$
- Absolute minimum: $-\sqrt{2}$ at $x = \frac{5\pi}{4}$

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.