To solve by completing the square, what needs to be moved in this equation? x² = 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

o-na [289]

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,

$f''(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 30x - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 30(x - 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow x - 2 \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{x \ \textgreater \ 2}$

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),

$f''(x) \ \textless \ 0 
\\ \\ \Leftrightarrow 30x - 60 \ \textless \ 0 
\\ \\ \Leftrightarrow 30(x - 2) \ \textless \ 0 
\\ \\ \Leftrightarrow x - 2 \ \textless \ 0 
\\ \\ \Leftrightarrow \boxed{x \ \textless \ 2}$

Therefore, f is concave down at $(-\infty, 2)$ .

3 0

3 years ago

The sum of two even consecutive integers is 46 find the intergers

Anarel [89]

Answer:

First number = 22

Second number = 24

Step-by-step explanation:

Let the first number = x

Let the second number = x + 2

According to the question ,

The sum of two even consecutive ingers = 46.

so,

x + x + 2 = 46

2x + 2 = 46

2x = 46 - 2

2x = 44

x = 44 / 2

x = 22

∴ FIRST NUMBER = X

= 22

SECOND NUMBER = X + 2

= 22 + 2

= 24.

4 0

3 years ago

Look at the system of equations below Sy= -2x +3 1 4x – 3y = 11

kompoz [17]

Answer:

(2,-1)

y= -1

x= 2

Step-by-step explanation:

y= -2x+3

4x-3y=11

substitute the value of y into an equation

4x-3(-2x+3)=11

distribute by multiplying numbers in parenthesis by -3

4x+6x-9=11

add 4x to 6x

10x-9=11

add 9 from both sides

10x=20

divide both sides by 10

x=2

substitute the value of x into an equation

y= -2•2+3

multiply -2 to 2

y= -4+3

add -4 to 3 but 4 is negative, so subtract

y= -1

-----------

(2,-1)

-----------

3 0

2 years ago

Jack is a professional weightlifter. his personal lifting record was 125.8kg. at a weightlifting competition, he broke his own r

Softa [21]

Answer:

126.34kg

Step-by-step explanation:

125.8+0.54=126.34kg

Convert 540 grams to kilograms` 0.54 kilograms, and add to the previous record

5 0

2 years ago

Find the equation of the line. Use exact numbers. y = x+

Alexxandr [17]

This is poorly written, but as i understand:

We have the general equation:

y = x + i

where i is a constant, and we want this function to pass through the point x = 2 and y = 3.

Then we must find the value of y, we know that when x= 2 we also must have y = 3, so we can replace those two values in the equation above and get

y = 3 = x + i = 2 + i

3 = 2 + i

3 - 2 = i

1 = i

So our equation is y = x + 1

8 0

2 years ago

To solve by completing the square, what needs to be moved in this equation? x² = 9 - 4x