[tex]f(x)=3\sin x+3\cos x\implies f'(x)=3\cos x-3\sin x[/tex]
[tex]f[/tex] has critical points where [tex]f'=0[/tex]:
[tex]3\cos x-3\sin x=0\implies\cos x=\sin x\implies\tan x=1\implies x=\pm\dfrac\pi4+2n\pi[/tex]
where [tex]n[/tex] is any integer. We get solutions in the interval [tex]\left[0,\frac\pi3\right][/tex] for [tex]n=0[/tex], for which [tex]x=\frac\pi4[/tex].
At this critical point, we have [tex]f\left(\frac\pi4\right)=3\sqrt2\approx4.243[/tex].
At the endpoints of the given interval, we have [tex]f(0)=3[/tex] and [tex]f\left(\frac\pi3\right)=\frac{3+3\sqrt3}2\approx4.098[/tex].
So we have the extreme values
[tex]\max\limits_{x\in\left[0,\frac\pi3\right]}f=3\sqrt2[/tex]
[tex]\min\limits_{x\in\left[0,\frac\pi3\right]}f=3[/tex]
The absolute minimum and maximum are located at [tex]x = 0[/tex] and [tex]x = \frac{\pi}{4}[/tex], respectively.
In this question we are going to apply first and second derivative tests to determine all minima and maxima in the function. Let be [tex]f(x) = 3\cdot \sin x + 3\cdot \cos x[/tex].
The first and second derivative tests are, respectively:
First derivative
[tex]f'(x) = 3\cdot \cos x -3\cdot \sin x[/tex] (1)
First derivative test
[tex]3\cdot \cos x - 3\cdot \sin x = 0[/tex]
[tex]\tan x = 1[/tex]
[tex]x = \tan^{-1} 1[/tex]
[tex]x = \frac{\pi}{4} + \pi\cdot i[/tex], [tex]i\in \mathbb{Z}[/tex]
Second derivative
[tex]f''(x) = -3\cdot \sin x -3\cdot \cos x[/tex] (2)
Second derivative test
There is an only value to be evaluated: [tex]x = \frac{\pi}{4}[/tex]
[tex]f''\left(\frac{\pi}{4} \right) = -3\cdot \sin \frac{\pi}{4} - 3\cdot \cos \frac{\pi}{4}[/tex]
[tex]f''\left(\frac{\pi}{4} \right) = -3\sqrt{2}[/tex]
Which means that [tex]x = \frac{\pi}{4}[/tex] represents an absolute maximum.
Lastly, we evaluate the function:
[tex]f\left(\frac{\pi}{4} \right) = 3\cdot \sin \frac{\pi}{4} + 3\cdot \cos \frac{\pi}{4}[/tex]
[tex]f\left(\frac{\pi}{4} \right) = 3\sqrt{2}[/tex]
Now we evaluate the function at each bound:
[tex]x = 0[/tex]
[tex]f(0) = 3\cdot \sin 0 + 3\cdot \cos 0[/tex]
[tex]f(0) = 3[/tex]
[tex]x = \frac{\pi}{3}[/tex]
[tex]f\left(\frac{\pi}{3} \right) = 3\cdot \sin \frac{\pi}{3} + 3\cdot \cos \frac{\pi}{3}[/tex]
[tex]f\left(\frac{\pi}{3} \right) = \frac{3+3\sqrt{2}}{2}[/tex]
The absolute minimum and maximum are located at [tex]x = 0[/tex] and [tex]x = \frac{\pi}{4}[/tex], respectively.
We kindly invite to check this question on maxima and minima: https://brainly.com/question/12870574
