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What is #f(x) = int 1/(x+3)-1/(x-2) dx# if #f(-1)=3 #?

Answer 1

Isaiah Allen

#f(x)=ln|3/2 (x+3)/(x-2)|+3#

#f(x) = int [1/(x+3)-1/(x-2)] dx = ln|x+3|-ln|x-2|+C# #qquad = ln|(x+3)/(x-2)|+C#

Since #f(-1) = 3#, we have

#3 = ln|(-1+3)/(-1-2)|+C = ln|-2/3|+C implies C = 3-ln|2/3|#

Thus

#f(x) = ln|(x+3)/(x-2)|+3-ln|2/3|=ln|3/2 (x+3)/(x-2)|+3#

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Answer 2

Elijah Abram

To find ( f(x) ), we integrate ( \frac{1}{x+3} - \frac{1}{x-2} ) with respect to ( x ). Then, given that ( f(-1) = 3 ), we can determine the value of the constant of integration.

( f(x) = \ln|x+3| - \ln|x-2| + C )

Given ( f(-1) = 3 ), we substitute ( x = -1 ) into the function:

( \ln|-1+3| - \ln|-1-2| + C = 3 )

( \ln|2| - \ln|-3| + C = 3 )

( \ln(2) - \ln(3) + C = 3 )

( \ln\left(\frac{2}{3}\right) + C = 3 )

( C = 3 - \ln\left(\frac{2}{3}\right) )

Thus, ( f(x) = \ln|x+3| - \ln|x-2| + 3 - \ln\left(\frac{2}{3}\right) )

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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.