The equation of the tangent line at x=1 can be written in point-slope form as
... L(x) = f'(1)(x -1) +f(1)
The derivative is ...
... f'(x) = 4x^3 +4x
so the slope of the tangent line is f'(1) = 4+4 = 8.
The value of the function at x=1 is
... f(1) = 1^4 +2·1^2 = 3
So, your linearization is ...
... L(x) = 8(x -1) +3
or
... L(x) = 8x -5
Answer:
The value of car after 5 years of exponentially depreciation is $16,213.36
Step-by-step explanation:
The initial amount of the car = i = $24,600
The depreciates rate of interest applied = r = 8%
Let The value of car after n years = A
The number of years = n = 5 years
Now,According to question
The value of car after n years = initial price of car ×
Or, The value of car after 5 years = i ×
Or, $A = $24,600 ×
Or, $A = $24,600 ×
Or, $A = $24,600 × 0.65908
Or, $A = $16,213.36
So, The value of car after 5 years = A = $16,213.36
Hence , The value of car after 5 years of exponentially depreciation is $16,213.36 . Answer