The average surcharge for non-account holders who use an ATM can be modeled asfix) = 0.004x3 - 0.061x20.299x0.899 dollarswhere x is the number of years since 1998, data between 1998 and 2007.t(a) Write the derivative model for f.f'(x)dollars per year is the rate of change of the average surcharge for non-account holders who use an ATM where x is the number of years since 1998. Data from 1998 to 2007.(b) Estimate the transaction fee in 2014. (Round your answer to two decimal places.)$(c) Estimate the rate of change of the ATM fee in 2012. (Round your answer to three decimal places.)per year$

Question
Asked Oct 18, 2019
The average surcharge for non-account holders who use an ATM can be modeled as
fix) = 0.004x3 - 0.061x2
0.299x0.899 dollars
where x is the number of years since 1998, data between 1998 and 2007.t
(a) Write the derivative model for f.
f'(x)
dollars per year is the rate of change of the average surcharge for non-account holders who use an ATM where x is the number of years since 1998. Data from 1998 to 2007.
(b) Estimate the transaction fee in 2014. (Round your answer to two decimal places.)
$
(c) Estimate the rate of change of the ATM fee in 2012. (Round your answer to three decimal places.)
per year
$
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The average surcharge for non-account holders who use an ATM can be modeled as fix) = 0.004x3 - 0.061x2 0.299x0.899 dollars where x is the number of years since 1998, data between 1998 and 2007.t (a) Write the derivative model for f. f'(x) dollars per year is the rate of change of the average surcharge for non-account holders who use an ATM where x is the number of years since 1998. Data from 1998 to 2007. (b) Estimate the transaction fee in 2014. (Round your answer to two decimal places.) $ (c) Estimate the rate of change of the ATM fee in 2012. (Round your answer to three decimal places.) per year $

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Step 1

(a) To find the derivative of the given function f as follows.

f(x) 0.004x3 - 0.061x2 0.299x +0.899
d
f(x)
d0.004x30.06 lx2 +0.299x +0.899
d
-(0.299x)+
d (0.899)
d
dr0.004x "(0.061x
dx
d
d
=0.004 r)-0.061-
dx
d
x)+0.299(x)+0
dx
=0.004(3x2)-0.061(2x) +0.299 (1)
= S(0.012x2-0.122.x + 0.299) per year
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f(x) 0.004x3 - 0.061x2 0.299x +0.899 d f(x) d0.004x30.06 lx2 +0.299x +0.899 d -(0.299x)+ d (0.899) d dr0.004x "(0.061x dx d d =0.004 r)-0.061- dx d x)+0.299(x)+0 dx =0.004(3x2)-0.061(2x) +0.299 (1) = S(0.012x2-0.122.x + 0.299) per year

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Step 2

(b) To find the estimate fee ...

No years from 1998 to 2014 = 16
f(x) 0.004x3-0.06 lx2 +0.299x+ 0.899
0.004(16)' - 0.061(16) +0.299(16) + 0.899
(16)
16.384-15.616+ 4.784 + 0.899
= $6.451
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No years from 1998 to 2014 = 16 f(x) 0.004x3-0.06 lx2 +0.299x+ 0.899 0.004(16)' - 0.061(16) +0.299(16) + 0.899 (16) 16.384-15.616+ 4.784 + 0.899 = $6.451

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Tagged in

Math

Calculus

Derivative