What is #f(x) = int (3-x)/(x-4) # if #f(5)=1 #?

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Answer 1 · 2017-01-23T03:47:33

#f(x) = 6 - x - ln|x- 4|#

#f(x) = int(-(x - 3))/(x- 4)dx#

#f(x) = -int(x - 3)/(x - 4)dx#

Decompose into partial fractions.

#A/1 + B/(x -4) = (x - 3)/(1(x - 4))#

#A(x- 4) + B(1) = x - 3#

#Ax - 4A + B = x - 3#

Write a system of equations:

#{(A = 1), (B - 4A = -3):}#

Solve:

#A = 1# and #B = 1#.

#f(x) =- int 1 + 1/(x - 4)dx#

This can be integrated using #int(1/x) = ln|x| + C# and #int(x^n)dx = x^(n +1)/(n + 1) + C#.

#f(x) = -x - ln|x- 4| + C#

Solve for #C# now.

#1 = -5 - ln|5 - 4| + C#

#1 = -5 + 0 + C#

#C = 6#

The function is therefore #f(x) = 6 - x - ln|x- 4|#.

Hopefully this helps!