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What is #f(x) = int 1/(x-4) # if #f(2)=1 #?

Answer 1

Paisley Johnson

#f(x)=ln|x-4|+ln(e/2)#

#f(x)=ln|x-4|+1-ln2#

Here,

#f(x)=int1/(x-4)dx#

#f(x)=ln|x-4|+c....to(1)#

Given that ,

#f(2)=1#

#=>ln|2-4|+c=1#

#=>ln|-2|+c=1#

#=>c=1-ln2#

#=>c=lne-ln2...to[becauselne=1]#

#=>c=ln(e/2)#

Subst. #c=ln(e/2)# , into #(1)#

#f(x)=ln|x-4|+ln(e/2)#

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Answer 2

Isabella Ackerman

To find the function ( f(x) ) when ( f(2) = 1 ) for the equation ( f(x) = \int \frac{1}{x - 4} ), you need to evaluate the definite integral.

The indefinite integral of ( \frac{1}{x - 4} ) is ( \ln|x - 4| + C ), where ( C ) is the constant of integration.

Using the given condition ( f(2) = 1 ), we can find the value of ( C ) by plugging in ( x = 2 ) and equating it to 1:

( \ln|2 - 4| + C = 1 )

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

( \ln 2 + C = 1 )

Now, solve for ( C ):

( C = 1 - \ln 2 )

Thus, the function ( f(x) ) is:

( f(x) = \ln|x - 4| + (1 - \ln 2) )

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Answer 3

Christopher Allers

The function f(x) = ∫(1/(x-4))dx, when f(2) = 1, would need to be evaluated by finding the definite integral of 1/(x-4) from the lower limit of integration to x = 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.