2 years ago

Hey can someone help me ?

1 answer:

6 0

Sure what do u need help with?

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Inga [223]

Answer: (-8,-5)

Step-by-step explanation:

6 0

2 years ago

Bond [772]

$f(x)=3\sin x+3\cos x\implies f'(x)=3\cos x-3\sin x$

$f$ has critical points where $f'=0$ :

$3\cos x-3\sin x=0\implies\cos x=\sin x\implies\tan x=1\implies x=\pm\dfrac\pi4+2n\pi$

where $n$ is any integer. We get solutions in the interval $\left[0,\frac\pi3\right]$ for $n=0$ , for which $x=\frac\pi4$ .

At this critical point, we have $f\left(\frac\pi4\right)=3\sqrt2\approx4.243$ .

At the endpoints of the given interval, we have $f(0)=3$ and $f\left(\frac\pi3\right)=\frac{3+3\sqrt3}2\approx4.098$ .

So we have the extreme values

$\max\limits_{x\in\left[0,\frac\pi3\right]}f=3\sqrt2$

$\min\limits_{x\in\left[0,\frac\pi3\right]}f=3$

8 0

3 years ago

WARRIOR [948]

27x^3-108x^2y+144xy^2-64y^3

5 0

3 years ago

jeka57 [31]

Answer:

noe lsoocslo oqiuepcn

Step-by-step explanation:

mcm

8 0

2 years ago

Olenka [21]

The answer is 60 it was easy

4 0

2 years ago