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What is #f(x) = int e^(-x-1)+2x dx# if #f(0) = -4 #?

Answer 1

Luke Alvarez

It is derived as follows:

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Answer 2

Cameron Alexander

To find ( f(x) ), integrate the function ( e^{-x-1} + 2x ) with respect to ( x ). Then use the given initial condition ( f(0) = -4 ) to determine the constant of integration.

[ f(x) = \int e^{-x-1} + 2x , dx ]

First, integrate ( e^{-x-1} ) and ( 2x ) separately:

[ \int e^{-x-1} , dx = -e^{-x-1} + C_1 ] [ \int 2x , dx = x^2 + C_2 ]

Combine the results:

[ f(x) = -e^{-x-1} + x^2 + C ]

Using the initial condition ( f(0) = -4 ), we have:

[ -1 + 0 + C = -4 ]

Thus, ( C = -3 ).

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

[ f(x) = -e^{-x-1} + x^2 - 3 ]

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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.