Question 1195931: This is a calculus question. Can you explain how to solve #12 on https://www.math.purdue.edu/php-scripts/courses/oldexams/serve_file.php?file=16200FE-F2018.pdf ?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Answer: E) -5/64
============================================================================
Explanation:
Review this page to get a refresher of what a Maclaurin Series is
https://mathworld.wolfram.com/MaclaurinSeries.html
It's a special type of Taylor Polynomial centered at x = 0
Let's break the original function up like so
f(x) = (x^2+1)/(x-2)
f(x) = g(x)/h(x)
g(x) = x^2+1
h(x) = x-2
Which means,
g ' (x) = 2x
h ' (x) = 1
Use the quotient rule to find the following:
f(x) = g(x)/h(x)
f ' (x) = (g ' (x)*h(x) - g(x)*h ' (x))/( h(x)^2 )
f ' (x) = (2x(x-2) - (x^2+1)(1))/( (x-2)^2 )
f ' (x) = (2x^2-4x - x^2-1)/( (x-2)^2 )
f ' (x) = (x^2-4x -1 )/( (x-2)^2 )
Apply the quotient rule again
f '' (x) = 2nd derivative
f '' (x) = d/dx[ f ' (x) ]
f '' (x) = d/dx[ (x^2-4x -1 )/( (x-2)^2 ) ]
f '' (x) = 10/( (x-2)^3 )
I skipped a bunch of steps, so let me know if you need to see them.
Also,
f ''' (x) = 3rd derivative of f(x)
f ''' (x) = d/dx[ f '' (x) ]
f ''' (x) = d/dx[ 10/( (x-2)^3 ) ]
f ''' (x) = -30/( (x-2)^4 )
You don't need the quotient rule this time.
Think of 10/( (x-2)^3 ) as 10(x-2)^(-3). Then apply the power rule and chain rule
And,
f4(x) = 4th derivative of f(x)
f4(x) = d/dx[ f ''' (x) ] ....
f4(x) = d/dx[ -30/( (x-2)^4 ) ]
f4(x) = 120/( (x-2)^5 )
You won't need the quotient rule here either. Follow similar steps as in the previous section.
And finally,
f5(x) = 5th derivative of f(x)
f5(x) = d/dx[ f4(x) ]
f5(x) = d/dx[ 120/( (x-2)^5 ) ]
f5(x) = -600/( (x-2)^6 )
f5(0) = -600/( (0-2)^6 )
f5(0) = -75/8
Like the previous two subsections, you won't need the quotient rule either.
Unlike the previous sections, I have evaluated at x = 0 since this output will help determine the coefficient of the x^5 term
The nth derivative term is part of the coefficient for x^n
That coefficient is (f5(0))/(5!) = (-75/8)/(120) = -5/64 which is why Choice E is the answer
You can use a calculator like this one
https://calculator-online.net/taylor-series-calculator/
to confirm the answer
Another way to verify the answer is to use the TaylorPolynomial function in GeoGebra.
You can use the regular input box, or the CAS input. I prefer the CAS for this case since it will keep the fraction coefficients intact.
f(x) = original function in blue
T(x) = taylor polynomial in red
Notice that the red curve T(x) is a somewhat close match to the blue curve f(x), but only around x = 0.
Otherwise, the error gets worse and worse the further you move away from x = 0.
Further Reading:
https://tutorial.math.lamar.edu/Classes/CalcII/TaylorSeries.aspx
The article talks about Taylor Series, but also mentions Maclaurin Series as well.
|
|
|
