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sladkih [1.3K]
3 years ago
13

Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. (Enter your answers using inte

rval notation. If the answer cannot be expressed as an interval, enter EMPTY or ∅.) f(x) = 6 − x2 x

Mathematics
1 answer:
Mashcka [7]3 years ago
8 0

Answer:

f'(x) > 0 on (-\infty ,\frac{1}{e}) and f'(x)<0 on(\frac{1}{e},\infty)

Step-by-step explanation:

1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:

f'(x)>0

To find its decreasing interval :

f'(x)

2) Then let's find the critical point of this function:

f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0

2.2 Solving for x this equation, this will lead us to one critical point since x'  is not defined for Real set, and x''=\frac{1}{e}≈0.37 for e≈2.72

x^{2x}=0 \ni \mathbb{R}\\(ln(x)+1)=0\Rightarrow ln(x)=-1\Rightarrow x=e^{-1}\Rightarrow x\approx 2.72^{-1}x\approx 0.37

3) Finally, check it out the critical point, i.e. f'(x) >0 and below f'(x)<0.

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Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

The answer should be A which is:
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7 0
3 years ago
A triangle has three different integer side lengths and a perimeter of 20 units. What is the max length of any one side?
Troyanec [42]
X + x+1 + x + 2 = 20
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3 years ago
You bicycle along a straight flat road with a safety light attached to one foot. Your bike moves at a speed of 15 km/hr and your
Viefleur [7K]

Answer:

a)

x(t) = ( 4.167 t + 0.2 cos (2πt))

y(t) = ( 0.3 - 0.2 sin  (2πt))

b) the  foot have to be 3.32 rev/sec faster

Step-by-step explanation:

Given that:

the speed of the bike = 15 km/hr = 15 × 1000/3600 (m/sec) = 4.167 m/sec

radius of the circle when the foot moves = 20 cm = 0.2 m

radius of the circle above the ground = 30 cm = 0.3 m

Let assume that:

x(t)  should represent the vector along the horizontal moment

y(t) should be the vector along the vertical moment

The initial component will be ( 0, 0.3)

We know that the radius of the circle is given as 0.2 m, So the vector of the circle can be written as (0.2 cos t , 0.2 sin t )

Also, the foot makes one revolution in a second, definitely the frequency of the revolution = 1 and the vector for the circle is ( 0.2 cos (2πt), -0.2 sin  (2πt)), due to the fact that the foot moves clockwise.

Thus, adding all the component together ; we have:

(x(t), y(t)) = (0,0.3)+(4.167 t , 0)+(0.2 cos (2πt), -0.2 sin  (2πt))

(x(t), y(t)) = (4.167 t + 0.2 cos (2πt), 0.3 - 0.2 sin  (2πt))

Hence; the parametric equations are:

x(t) = ( 4.167 t + 0.2 cos (2πt))

y(t) = ( 0.3 - 0.2 sin  (2πt))

b)

The linear speed of rotation is :

15km/hr = 15 × 100, 000/3600 (cm/sec)

             = 416.7 cm/sec

The rotational frequency is :

= 416.7/2πr

= 416.7/2(3.14 × 20)

= 3.32 rev/sec

Hence, the  foot have to be 3.32 rev/sec faster in rotating if an observer standing at the side of the road sees the light moving backward.

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2 years ago
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Answer:

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Step-by-step explanation:

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