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IrinaVladis [17]
3 years ago
9

Let f(x)=5x3−60x+5 input the interval(s) on which f is increasing. (-inf,-2)u(2,inf) input the interval(s) on which f is decreas

ing. (-2,2) find the intervals on which f is concave up. (-inf,-2)u(2,inf) find the intervals on which f is concave down.
Mathematics
1 answer:
o-na [289]3 years ago
3 0
Answers:

(a) f is increasing at (-\infty,-2) \cup (2,\infty).

(b) f is decreasing at (-2,2).

(c) f is concave up at (2, \infty)

(d) f is concave down at (-\infty, 2)

Explanations:

(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that

f'(x) = 15x^2 - 60


So,


f'(x) \ \textgreater \  0
\\
\\ \Leftrightarrow 15x^2 - 60 \ \textgreater \  0
\\
\\ \Leftrightarrow 15(x - 2)(x + 2) \ \textgreater \  0
\\
\\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textgreater \  0} \text{   (1)}

The zeroes of (x - 2)(x + 2) are 2 and -2. So we can obtain sign of (x - 2)(x + 2) by considering the following possible values of x:

-->> x < -2
-->> -2 < x < 2
--->> x > 2

If x < -2, then (x - 2) and (x + 2) are both negative. Thus, (x - 2)(x + 2) > 0.

If -2 < x < 2, then x + 2 is positive but x - 2 is negative. So, (x - 2)(x + 2) < 0.
 If x > 2, then (x - 2) and (x + 2) are both positive. Thus, (x - 2)(x + 2) > 0.

So, (x - 2)(x + 2) is positive when x < -2 or x > 2. Since

f'(x) \ \textgreater \  0 \Leftrightarrow (x - 2)(x + 2)  \ \textgreater \  0

Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at (-\infty,-2) \cup (2,\infty).

(b) f is decreasing only when the derivative of f is negative. Since

f'(x) = 15x^2 - 60

Using the similar computation in (a), 

f'(x) \ \textless \  \ 0 \\ \\ \Leftrightarrow 15x^2 - 60 \ \textless \  0 \\ \\ \Leftrightarrow 15(x - 2)(x + 2) \ \ \textless \  0 \\ \\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textless \  0} \text{ (2)}

Based on the computation in (a), (x - 2)(x + 2) < 0 only when -2 < x < 2.

Thus, f'(x) < 0 if and only if -2 < x < 2. Hence f is decreasing at (-2, 2)

(c) f is concave up if and only if the second derivative of f is positive. Note that

f''(x) = 30x - 60

Since,

f''(x) \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow 30x - 60 \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow 30(x - 2) \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow x - 2 \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow \boxed{x \ \textgreater \  2}

Therefore, f is concave up at (2, \infty).

(d) Note that f is concave down if and only if the second derivative of f is negative. Since,

f''(x) = 30x - 60

Using the similar computation in (c), 

f''(x) \ \textless \  0 &#10;\\ \\ \Leftrightarrow 30x - 60 \ \textless \  0 &#10;\\ \\ \Leftrightarrow 30(x - 2) \ \textless \  0 &#10;\\ \\ \Leftrightarrow x - 2 \ \textless \  0 &#10;\\ \\ \Leftrightarrow \boxed{x \ \textless \  2}

Therefore, f is concave down at (-\infty, 2).
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raketka [301]

Answer:

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7 0
2 years ago
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8xy= 2(2x + y) solve for y
STatiana [176]

Answer:  \displaystyle y= \frac{2\text{x}}{4\text{x} - 1}

============================================

Work Shown:

8\text{x}y= 2(2\text{x} + y)\\\\8\text{x}y= 4\text{x} + 2y\\\\8\text{x}y - 2y= 4\text{x}\\\\y(8\text{x} - 2)= 4\text{x}\\\\y= \frac{4\text{x}}{8\text{x} - 2}\\\\y= \frac{2*2\text{x}}{2(4\text{x} - 1)}\\\\y= \frac{2\text{x}}{4\text{x} - 1}\\\\

---------------

Explanation:

The idea is to first get all the y terms to one side. I did this by subtracting 2y from both sides in the third step.

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Afterward there's the optional steps of simplifying (as shown in the final two steps).

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Mr. Clifton was preparing for his upcoming birthday party. He ordered 51 white balloons and 47 red balloons. As he walked to his
Andrej [43]

Answer:

28 i think

Step-by-step explanation:

51+47=98

98-12 the ones that flew away that takes it to 86

86/3 the 3 balloons buntch than is 28

p.s. i mght of got some equations wrong but this is what i got

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3 years ago
An opaque cylindrical tank with an open top has a diameter of 3.25 m and is completely filled with water. When the afternoon Sun
kicyunya [14]

Answer:

10.69

Step-by-step explanation:

We must first know that the angle of refraction is given by the following equation:

n_water * sin (A_water) = n_air * sin (A_air)

where n is the refractive index, for water it is 1.33 and for air it is 1.

The angle (A) in the air is 22.8 °, and that of the water is unknown.

Replacing these values we have to:

1.33 * sin (A_water) = 1 * 0.387

sin (A_water) = 1 * 0.387 / 1.33

A_water = arc sin (0.2907) = 16.9 °

now for the tank depth:

h = D / tan (A_water)

D = 3.25

Replacing

h = 3.25 / tan 16.9 °

h = 10.69

Therefore the depth is 10.69 meters.

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3 years ago
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