Find a function f(x) such that: f'(x) = x^(3)and the line x + y = 0 is tangent to the graph of of f(x).  Use U-substitution to get answers

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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`f'(x) = x^3`

 

`f(x) = int x^3dx`

 

Let;

`U = x^4/4`

`dU = 4x^3/4 dx = x^3 dx`

 

So;

f(x)

`= int dU`

`= U+C`

`= x^4/4+C`

C is a constant.

 

`f(x) = x^4/4+C`

 


For any graph f(x) the gradient of the tangent line to f(x) at any point will be given by f'(x).

`f'(x) = x^3`

It is given that x+y = 0 is a tangent.

y = -x

So the gradient is -1.

`f'(x) = -1`

`x^3 = -1`

    `x = -1`

So at x=-1 there is a tangent to the graph.

This point will satisfy both f(x) and x+y = 0 at x= -1

`x^4/4+C = -x`

`(-1)^4/4+C = 1`

` C = 1-1/4`

   `= 3/4`

 

So;

`f(x) `

`= x^4/4+3/4 `

`= 1/4(x^4+3)`

 

`f(x) = 1/4(x^4+3)`

 

 

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