hectictar

a)

f(x)   =   3x - 2       Instead of  " f(x) "  let's say  y .

y  =  3x - 2            Now let's solve this equation for  x .

                             Add  2  to both sides.

y + 2  =  3x

                             Divide both sides by  3 .

(y + 2)/3  =  x

                             So the inverse function is....

f^-1(x)  =  (x + 2)/3

b)   I guess that this means find  (g o f)(x) .

(g o f)(x)  =  g( f(x) )

(g o f)(x)  =  g( 3x - 2 )

(g o f)(x)  =  10/( 3x - 2 ) + 2

\(x = {-2m \pm \sqrt{(2m)^2-4(m^2+2m+3)} \over 2} = {-2m \pm \sqrt{4m^2-4m^2-8m-12} \over 2}= -m \pm \sqrt{-2m-3} \)

\(r = -m + \sqrt{-2m-3} \\~\\ r^2=(-m + \sqrt{-2m-3})(-m + \sqrt{-2m-3}) \\~\\ r^2=m^2-2m\sqrt{-2m-3}+(-2m-3)\)

and

\(s = -m-\sqrt{-2m-3} \\~\\ s^2=(-m-\sqrt{-2m-3})(-m-\sqrt{-2m-3}) \\~\\ s^2=m^2+2m\sqrt{-2m-3}+(-2m-3)\)

\(r^2+s^2=m^2-2m\sqrt{-2m-3}+(-2m-3)+m^2+2m\sqrt{-2m-3}+(-2m-3) \\~\\ r^2+s^2=2m^2+2(-2m-3) \\~\\ r^2+s^2=2m^2-4m-6\)

The minimum value of  r^2 + s^2  occurs when  m   =   4/( 2 * 2 )   =   1

When m = 1 ,   r² + s²   =   2 - 4 - 6   =   -8

1.

\(y=\frac{20}{1+9e^{3x}} \\~\\ y(1 + 9e^{3x})=20 \\~\\ 1 + 9e^{3x}=\frac{20}{y} \\~\\ 9e^{3x}=\frac{20}{y}-1 \\~\\ 9e^{3x}=\frac{20-y}{y} \\~\\ e^{3x}=\frac{20-y}{9y} \\~\\ 3x=\ln(\frac{20-y}{9y}) \\~\\ x=\frac{\ln(\frac{20-y}{9y})}{3}\)

There is an asymptote when   9y  =  0   which is when   y = 0

There is an asymptote when   \(\frac{20-y}{9y}\)  =  0   which is when   y = 20

In this case  5 - 4  =  1  so the denominator just equals 1 

\(\lim\limits_{ x \to 25^+ }f(x)\)  must be equal to \(\lim\limits_{ x \to 25^- }f(x)\)  so that the limit exists at the point with x-coordinate 25.

(4/x+2)/(1/2x-8)   =   \(\frac{\frac4x+2}{\frac12x-8}\)

Only the first number after the division sign goes in the denominator when there aren't parenthesees.

1/2x   means   1/2 * x   which is   \(\frac12\) * x

and

1/(2x)   means   \(\frac1{2x}\) 

Let  J  be the number of miles between Janet and the office.

Let  L  be the number of miles between Lynn and the office.

Janet and Lynn live 8 mi apart in opposite directions from their office.

J + L  =  8

Lynn lives 1 mi less than twice as far from the office as Janet does.

L  =  2J - 1    Use this value for  L  in the first equation.

J + (2J - 1)  =  8

3J - 1  =  8

3J  =  9

J  =  3     Use this value for  J  to find  L .

L  =  2J - 1

L  =  2(3) - 1

L  =  5

Janet lives  3  miles from the office, and Lynn lives  5  miles from the office.

Let  m = the number of men

Let  w = the number of women

There are 3 fewer men than twice the number of women.

m  =  2w - 3

If there were 2 more women on the jury, the numbers of men and women would be equal.

w + 2  =  m       Subtract  2  from both sides of this equation to solve for  w .

w  =  m - 2        Now we can use this value of  w  in the first equation.

m  =  2w - 3

m  =  2(m - 2) - 3

m  =  2m - 4 - 3

m  =  2m - 7

m - 2m  =  -7

-m  =  -7

m  =  7  =  the number of men

Gracias

volume of candle   =   \(\frac13\pi r^2h\)          Since  h = r , we can replace  h  with  r .

volume of candle   =   \(\frac13\pi r^2r\)

volume of candle   =   \(\frac13\pi r^3\)

surface area of candle   =   \(\pi r^2+\pi r\sqrt{r^2+h^2}\)        Since  h = r , we can replace  h  with  r .

surface area of candle   =   \(\pi r^2+\pi r\sqrt{r^2+r^2}\)

surface area of candle   =   \(\pi r^2+\pi r\sqrt{2r^2}\)

surface area of candle   =   \(\pi r^2+\pi r\cdot r\sqrt{2}\)

surface area of candle   =   \(\pi r^2+\pi r^2\sqrt{2}\)

ratio of the volume of candle to its surface area  =   \(\frac{\text{volume of candle}}{\text{surface area of candle}}\)

\(\frac{\text{volume of candle}}{\text{surface area of candle}}\,=\,\frac{\frac13\pi r^3}{\pi r^2+\pi r^2\sqrt2}\\~\\ =\,\frac{\frac13r^3}{r^2+r^2\sqrt2} \\~\\ =\,\frac{\frac13r}{1+\sqrt2} \\~\\ =\,\frac{r}{3(1+\sqrt2)} \\~\\ =\,\frac{r(1-\sqrt2)}{3(1+\sqrt2)(1-\sqrt2)} \\~\\ =\,\frac{r(1-\sqrt2)}{3(1-2)} \\~\\ =\,\frac{r(1-\sqrt2)}{-3} \)