How do you find a function f(x), which, when multiplied by its derivative, gives you #x^3#, and for which #f(0) = 4#?

Answer 1
The answer is: #y=sqrt(x^4/2+16)#.

This is a first-order differential equation with separable variables:

#yy'=x^3rArry(dy)/(dx)=x^3rArrydy=x^3dxrArr#
#intydy=intx^3dxrArry^2/2=x^4/4+crArry=+-sqrt(x^4/2+2c)#

Given that the starting circumstance is:

#f(0)=4#, we have to choose the positive one, and than we can find the constant #c#:
#4=sqrt(0/2+2c)rArr16=2crArrc=8#.

So the function is:

#y=sqrt(x^4/2+16)#.
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Answer 2

To find the function ( f(x) ), we need to solve the differential equation

[ f(x) \cdot \frac{{df}}{{dx}} = x^3 ]

Given that ( f(0) = 4 ), we can proceed as follows:

  1. Separate variables and integrate both sides with respect to ( x ):

[ \int f(x) , df = \int x^3 , dx ]

  1. Integrate both sides:

[ \frac{1}{2} f(x)^2 = \frac{1}{4} x^4 + C ]

  1. Solve for ( f(x) ):

[ f(x)^2 = \frac{1}{2} x^4 + C ]

  1. Apply the initial condition ( f(0) = 4 ) to find the constant ( C ):

[ f(0)^2 = C ] [ 4^2 = C ] [ C = 16 ]

  1. Substitute ( C = 16 ) back into the equation:

[ f(x)^2 = \frac{1}{2} x^4 + 16 ]

  1. Take the square root of both sides:

[ f(x) = \pm \sqrt{\frac{1}{2} x^4 + 16} ]

Given ( f(0) = 4 ), we choose the positive square root:

[ f(x) = \sqrt{\frac{1}{2} x^4 + 16} ]

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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.

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.

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.

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.

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