# The graph of a function f is shown.?Use the differential equation and the given point to find an equation of the function dy/dx = 4x+ (9x^2) / ( (3x^3 +1) ^3/2 )...

The graph of a function f is shown.?

Use the differential equation and the given point to find an equation of the function

dy/dx = 4x+ (9x^2) / ( (3x^3 +1) ^3/2 )

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### 1 Answer

You need to find the function using the process of integration such that:

`int dy = int (4x+ (9x^2)/((3x^3 +1)^(3/2))) dx`

Using the property of linearity of integral yields:

`y = int 4x dx + int (9x^2)/((3x^3 +1)^(3/2))) dx`

`y = 4x^2/2 + int (9x^2)/((3x^3 +1)^(3/2))) dx`

You need to solve the integral `int (9x^2)/((3x^3 +1)^(3/2))) dx ` using the following substitution such that:

`3x^3 + 1 = t => 9x^2 dx = dt`

Changing the variable yields:

`int (9x^2)/((3x^3 +1)^(3/2))) dx= int (dt)/(t^(3/2))`

Converting the power `1/(t^(3/2))` into a negative power yields:

`int (dt)/(t^(3/2)) = int (t^(-3/2)) dt`

`int (t^(-3/2)) dt = (t^(-3/2+1))/(-3/2+1) + c`

`int (t^(-3/2)) dt = -2/sqrt t + c`

Substituting back `3x^3 + 1` for t yields:

`int (9x^2)/((3x^3 +1)^(3/2))) dx = -2/sqrt(3x^3 + 1) + c`

`y = 4x^2/2 - 2/sqrt (3x^3 + 1) + c`

`y = 2x^2 - 2/sqrt (3x^3 + 1) + c`

The problem provides the information that the graph of f(x)=y intercepts y axis at `(0,2), ` hence, you may use this information to find the constant c such that:

`f(0) = 2*0^2 - 2/sqrt (3*0^3 + 1) + c`

`2 = -2 + c => c = 4`

`y = 2x^2 - 2/sqrt (3x^3 + 1) + 4`

**Hence, evaluating the function f(x)=y, using the given conditions, yields `y = 2x^2 - 2/sqrt (3x^3 + 1) + 4.` **