What is f(x) if f'(x) = 2x/x^2-1 and f(2) = 0?
If
f'(x) = 2x/x^2-1
mean
f'(x) = 2 x / ( x² -1 )
then
f (x) = ∫ f'(x) dx = ∫ 2 x dx / ( x² -1 )
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Remark:
Substitution
x² -1 = u
2 x dx = du
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f (x) = ∫ du / u = ℓn ( u ) + C
f (x) = ℓn ( x² -1 ) + C
f(2) = 0
f (2) = ℓn ( 2² -1 ) + C = 0
ℓn ( 4 -1 ) + C = 0
ℓn ( 3 ) + C = 0
Subtract ℓn ( 3 ) to both sides
ℓn ( 3 ) + C - ℓn ( 3 ) = 0 - ℓn ( 3 )
C = - ℓn ( 3 )
f (x) = ℓn ( x² -1 ) + C
f (x) = ℓn ( x² -1 ) - ℓn ( 3 )
f (x) = ℓn [ ( x² -1 ) / 3 ]
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Remark:
ℓn ( a ) - ℓn ( b) = ℓn ( a / b )
so
ℓn ( x² -1 ) - ℓn ( 3 ) = ℓn [ ( x² -1 ) / 3 ]
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ℓn mean natural logarithm ( logarithm to the base e )
Thank you, Bosnian brother.
That is exactly what I put on my test, I was just making sure.
To find the function f(x), we first need to integrate the derivative f'(x) to obtain f(x).
Given: f'(x) = 2x/(x^2 - 1)
To integrate f'(x), we can rewrite the expression as:
f'(x) = 2x * (x^2 - 1)^(-1)
Now, let's integrate f'(x) with respect to x:
∫ f'(x) dx = ∫ 2x * (x^2 - 1)^(-1) dx
To integrate this, we can use a substitution. Let u = x^2 - 1, then du = 2x dx. Substituting these values, we have:
∫ f'(x) dx = ∫ (1/u) du
Now we integrate ∫ (1/u) du as:
∫ (1/u) du = ln|u| + C
where C is the constant of integration.
Substituting back u = x^2 - 1, we obtain:
∫ f'(x) dx = ∫ (1/(x^2 - 1)) dx = ln|x^2 - 1| + C
Now, to find the constant C, we use the given information f(2) = 0.
Substitute x = 2 into the equation:
ln|2^2 - 1| + C = ln|4 - 1| + C = ln|3| + C
Since f(2) = 0, we know that ln|3| + C = 0. Solving for C, we have:
C = -ln|3|
Finally, substituting the value of C into our equation, we have:
∫ f'(x) dx = ln|x^2 - 1| - ln|3|
Therefore, the function f(x) is given by:
f(x) = ln|x^2 - 1| - ln|3|