exists and is bounded for all 
. We're told that 
. Consider the interval [0, 3]. The mean value theorem says that there is some 
such that
Since 
, we have
so 19 is the largest possible value.
Answer:
x = 245
Step-by-step explanation:
The sum of integers x and y is 280; x + y = 280
If x is divided by 7, the quotient is y; x/7 = y so x = 7y
Now you have 2 equations:
x + y = 280
x = 7y
Substitute x = 7y into x + y = 280
7y + y = 280
8y = 280
y = 35
x = 7(35)
x = 245
Answer
x = 245
For this case we have the following function:
f (x) = x ^ 5
We apply the following transformations:
reflect over the y axis:
f (x) = (- x) ^ 5
shift 1 unit left:
f (x) = (- x + 1) ^ 5
vertically compress by 1/3:
f (x) = (1/3) (- x + 1) ^ 5
Answer:
A function that results after applying the sequence of transformations is:
f (x) = (1/3) (- x + 1) ^ 5
Apply the rule: 
![3[2 ln(x-1) - lnx] + ln(x+1)=3[ln(x-1)^{2} - lnx ] + ln(x+1)](https://tex.z-dn.net/?f=3%5B2%20ln%28x-1%29%20-%20lnx%5D%20%2B%20ln%28x%2B1%29%3D3%5Bln%28x-1%29%5E%7B2%7D%20-%20lnx%20%5D%20%2B%20ln%28x%2B1%29)
Apply the rule : 
![3[2 ln(x-1) - lnx] + ln(x+1)=3ln\frac{(x-1)^{2} }{x} + ln(x+1)](https://tex.z-dn.net/?f=3%5B2%20ln%28x-1%29%20-%20lnx%5D%20%2B%20ln%28x%2B1%29%3D3ln%5Cfrac%7B%28x-1%29%5E%7B2%7D%20%7D%7Bx%7D%20%2B%20ln%28x%2B1%29)
Apply the rule:
![3[ln (x-1)^{2} -ln x]+ln (x+1)= ln \frac{(x-1)^{6} }{x^{3} } +log(x+1)](https://tex.z-dn.net/?f=3%5Bln%20%28x-1%29%5E%7B2%7D%20-ln%20x%5D%2Bln%20%28x%2B1%29%3D%20ln%20%5Cfrac%7B%28x-1%29%5E%7B6%7D%20%7D%7Bx%5E%7B3%7D%20%7D%20%2Blog%28x%2B1%29)
Finally, apply the rule: log a + log b = log ab
![3[ln(x-1)^{2} -ln x]+log(x+1)=ln\frac{(x-1)^{6}(x+1) }{x^{3} }](https://tex.z-dn.net/?f=3%5Bln%28x-1%29%5E%7B2%7D%20-ln%20x%5D%2Blog%28x%2B1%29%3Dln%5Cfrac%7B%28x-1%29%5E%7B6%7D%28x%2B1%29%20%7D%7Bx%5E%7B3%7D%20%7D)
To write the given quadratic equation to its vertex form, we first form a perfect square.
x² - 2x + 5 = 0
Transpose the constant to other side of the equation,
x² - 2x = -5
Complete the square in the left side of the equation,
x² - 2x + (-2/1(2))² = -5 + (-2/1(2))²
Performed the operation,
x² - 2x + 1 = -5 + 1
Factor the left side of the equation,
(x - 1)² = -4
Thus, the vertex form of the equation is,
<em> (x-1)² + 4 = 0</em>
