**** Please do all of the questions, i have several people skip these specific o
ID: 2882775 • Letter: #
Question
**** Please do all of the questions, i have several people skip these specific ones on previous attempts will rate 5 stars with answer***
Suppose that we have two functions, f(x) and g (x), and that
f (5) = 5,
g (5) = 5,
f '(5) = -2, and
g '(5) = 4.
Calculate the values of the following derivatives when x is equal to 5.
2)The number of thousand mp3 players produced per month by a certain manufacturer t months after the start of the year 2004 can be described by the function
As they sell more players, the company's profit increases both because they are selling more players and because economies of scale mean they can produce each player more cheaply.
The manufacturer's total profit (in thousands of dollars) when x thousand mp3 players are made is given by the function
We assume that the company sells all of the mp3 players that it produces.
Calculate the rate at which the sales of mp3 players increases (in thousands of players per month) in terms of t.
Rate of increase : ===???? thousand players per month.
Calculate the rate at which the manufacturer's profit increases (in thousands of dollars per thousand mp3 players) in terms of x.
Rate of increase : ==??? thousand dollars per thousand mp3 players.
Calculate the rate at which the manufacturer's profit increases at time t (in thousands of dollars per month).
Leave your answer in terms of t alone (x should not appear in your answer).
Rate of increase : ===??? thousand dollars per month.
3)An ice cube is melting, and the lengths of its sides are decreasing at a rate of 0.3 millimeters per minute.
At what rate is the volume of the ice cube decreasing when the lengths of the sides of the cube are equal to 11 millimeters?
Give your answer correct to the nearest cubic millimeter per minute.
Rate of decrease : === ???millimeters3 per minute.
Explanation / Answer
Solution:
f (5) = 5,
g (5) = 5,
f '(5) = -2, and
g '(5) = 4.
f(x)+g(x) = 5+5=10
f(x)g(x) =5*5=25
(f(x))/(g(x)) =5/5 = 1
f(x)/((f(x)+g(x)) = 5/(5+5) = 1/2
((f(x)g(x))^2) = (5*5)2 = 100
x(f(x)) = 5*5 = 25
f(x)/x = 5/5 = 1
(OR)
(f+g)' = f '+g' ==> -2 + 4 = 2
(fg)' = f 'g+fg' ==> -2*5 + 2*4 = 2
(f/g)' = (f 'g-fg')/g^2 ==> (-2*5 - 5*4)/52 = (-10 - 20)/25 = -6/5
now replacce g by h in then above formula and let h = (f+g) so
(f/h)' = (f 'h-fh')/h2
h'=f ' + g'
=> (f/(f+g))' = (f '(f+g) - f(f ' +g'))/(f+g)2 = (-2*(5+5)-5*(-2+4))/(5+5)2
= (-20-10)/100
= -30/100= -3/10
((fg)2)' = ((fg)*(fg))' = (fg)'fg + fg*(fg)' = 2*fg(fg)' = 2*fg*(fg'+f ' g)
= 2*5*5*(5*4-2*5)
= 50*(20-10) = 50*(10) = 500
(xn)' = nx(n-1) with n=1 we get x' = 1*1=1 (value of x doesn't matter)
x alone is equivalent to i(x)=x so i'(x) = 1
finally :
(xf)' = (if)' = f ' i + fi' = xf ' + f = 5*(-2) +5 = -10 + 5 = -5
now we are warm and
(f(x)/x)' = (f/i)' =(f ' i - fi')/i2 = (-2*2 - 5*1)/12 = -4 - 5 = -9
2)
df/dt = 0.4t + 5
dg/dx = 0.02x + 4
dg/dt = dg/dx*dx/dt
dx/dt = df/dt
dg/dt = (0.02x + 4)(0.4t + 5)
dg/dt = (0.02*(0.2t2 + 5t + 6) + 4)(0.4t + 5)
dg/dt = (0.004t2 + 0.1t + 0.12)(0.4t + 5)
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