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Find the y-intercept (the value of y when x=0) of the curve y=f(x) that goes though the...

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maheen100 | Student, Undergraduate | Honors

Posted March 2, 2013 at 6:29 PM via web

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Find the y-intercept (the value of y when x=0) of the curve y=f(x) that goes though the point (1,0) such that dy/dx=6xe^(-x^2)

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted March 2, 2013 at 6:43 PM (Answer #1)

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You need to find the function `f(x)` , such that:

`(dy)/(dx) = 6x*e^(-x^2)`

Separating the variables, yields:

`dy = 6x*e^(-x^2) dx`

Integrating both sides, yields:

int dy = int 6x*e^(-x^2) dx

`y = 6 int x*e^(-x^2) dx`

You should come up with the following substitution, such that:

`-x^2 = t => -2xdx = dt => xdx = -(dt)/2`

Changing the variable, yields:

`y = -6/2 int e^t dt => y = -3 e^t + c`

Substituting back `-x^2` for `t` yields:

`y = -3 e^(-x^2) + c`

Considering x = 1 and y = 0 yields:

`0 = -3*e^-1 + c => 1 = -3/e + c => c = 1.1

`Hence, evaluating the function `f(x) = y` yields `f(x) = -3 e^(-x^2) + 1.1` and the curve f(x) intercepts y axis at `y = 1.1` .

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