Can someone help me!
3. The value of a car (in $), 𝑥𝑥years after it is purchased, is given by:
v(x) = 32000/(((x/5)+1)^2)
a) What is the average annual price change for the first 5 years?
b) What is the instantaneous rate of change of 𝑉 when 𝑥 = 5 ?
v(x) = 32000/ (x/5 + 1)^2
average annual price change for the first 5 years
= (v(5) - v(0) )/(5-0)
= 32000/ (x/5 + 1)^2)
= (32000 / (1+1)^2 - 32000/(0+1)^1 )/5
= (8000-32000)/ 5 = -4800 , the negative shows me the value is decreasing
b) v ' (x) = -64000(x/5 + 1)^-3 (1/5)
= -12800/(x/5 + 1)^3
v ' (5) = -12800/(1+1)^3 = - 1600
v(x) = 32,000 / { (x/5)+1}^2
v(0) = 32,000
v(5) = 32,000/ { (5/5)+1}^2 = 32,000 / 4 = 8,000
change over five = 24,000
average = 24,000 / 5 = 4,800/year
dV/dx = -32000 * 2 { (x/5)+1}{1/5} / { (x/5)+1}^4
when x = 5
dV/dx = 6400 {2}{1/5}/ 8 =320
a) To find the average annual price change for the first 5 years, we need to find the difference in the value of the car after 5 years and its initial value, and then divide it by 5.
1. Substitute x = 5 into the given equation for v(x):
v(5) = 32000/(((5/5)+1)^2)
v(5) = 32000/((1+1)^2)
v(5) = 32000/(2^2)
v(5) = 32000/4
v(5) = 8000
2. Substitute x = 0 into the given equation for v(x) to find the initial value of the car:
v(0) = 32000/(((0/5)+1)^2)
v(0) = 32000/((0+1)^2)
v(0) = 32000/(1^2)
v(0) = 32000/1
v(0) = 32000
3. Calculate the difference in the values after 5 years and the initial value:
8000 - 32000 = -24000
4. Divide the difference by 5 to find the average annual price change:
-24000 / 5 = -4800
Therefore, the average annual price change for the first 5 years is -$4800 (the negative sign indicates a decrease in value).
b) To find the instantaneous rate of change of V when x = 5, we need to find the derivative of the given equation, v(x), with respect to x and then substitute x = 5 into the derivative equation.
To differentiate v(x), we'll use the quotient rule. The quotient rule states that if we have the function f(x) = g(x)/h(x), then the derivative of f(x) is given by:
[f'(x)]= [(g'(x) * h(x)) - (g(x) * h'(x))]/[h(x)^2]
Where f'(x) denotes the derivative of f with respect to x, g'(x) denotes the derivative of g with respect to x, and h'(x) denotes the derivative of h with respect to x.
1. Differentiate the numerator of v(x) = 32000/(((x/5)+1)^2):
[g(x)] = 32000
[g'(x)] = 0 (since constant)
2. Differentiate the denominator of v(x) = 32000/(((x/5)+1)^2):
[h(x)] = (((x/5)+1)^2)
[h'(x)] = 2 * ((x/5)+1) * (1/5)
[h'(x)] = (2/5) * (x/5 + 1)
3. Apply the quotient rule to calculate the derivative of v(x):
[v'(x)] = [(g'(x) * h(x)) - (g(x) * h'(x))]/[h(x)^2]
[v'(x)] = (0 * (((x/5)+1)^2)) - (32000 * ((2/5) * (x/5 + 1)))/((((x/5)+1)^2))^2
[v'(x)] = (-6400 * ((x/5)+1))/((((x/5)+1)^2))^2
4. Substitute x = 5 into the derivative equation, v'(x), to find the instantaneous rate of change:
[v'(5)] = (-6400 * ((5/5)+1))/((((5/5)+1)^2))^2
[v'(5)] = (-6400 * (2))/((2^2))^2
[v'(5)] = (-6400 * (2))/(4)^2
[v'(5)] = (-12800)/16
[v'(5)] = -800
Therefore, the instantaneous rate of change of V when x = 5 is -800.