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What is #f(x) = int (x-2)/(x-1) dx# if #f(2) = 0 #?

Answer 1

#f(x)=int(x-2)/(x-1)dx# #f(x)=int(x-1-1)/(x-1)dx# #f(x)=int(1+(-1)/(x-1))dx# #f(x)=int1dx-int(1)/(x-1)dx# #f(x)=x-log(x-1)+c# c is constant of integration to find the value of c we use #f(2)=0# If #f(2)=0# #=>2-log(2-1)+c=0# #=>2-log1+c=0# #=>2-0+c=0# #=>c=-2#

Hence #f(x)=x-log(x-1)-2#

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Answer 2

Paisley Johnson

Given ( f(2) = 0 ), we can find ( f(x) ) by finding the definite integral of ( \frac{x-2}{x-1} ) from some constant to ( x ), and then evaluating it at ( x = 2 ).

Let's find ( f(x) ) using the Fundamental Theorem of Calculus. The antiderivative of ( \frac{x-2}{x-1} ) is ( x - 2\ln|x-1| + C ).

So, ( f(x) = x - 2\ln|x-1| + C ).

Now, use ( f(2) = 0 ) to find ( C ):

[ f(2) = 2 - 2\ln|2-1| + C = 0 ] [ 2 - 2\ln|1| + C = 0 ] [ 2 - 2 \cdot 0 + C = 0 ] [ 2 + C = 0 ] [ C = -2 ]

Thus, ( f(x) = x - 2\ln|x-1| - 2 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.