How do you find an equation for the function #f'(x)=2x(4x^2-10)^2# whose graph passes through the point (2,10)?

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Answer 1 · 2017-01-19T16:00:10

The function is #y = 16/3x^6 - 40x^4 + 100x^2- 274/3#.

I would recommend expanding and then integrating.

#f'(x) = 2x(16x^4 - 80x^2 + 100)#

#f'(x) = 32x^5 - 160x^3 + 200x#

Write in Leibniz Notation:

#dy/dx = 32x^5 - 160x^3 + 200x#

We now have a separable differential equation.

#dy = (32x^5 - 160x^3 + 200x)dx#

#intdy = int(32x^5 - 160x^3 + 200x)dx#

#y = 16/3x^6 - 40x^4 + 100x^2 + C#

Solve for C now:

#10 = 16/3(2)^6 - 40(2)^4 + 100(2)^2 + C#

#10 = 1024/3 - 640 + 400 + C#

#250 - 1024/3 = C#

#C = -274/3#

The function is therefore #y = 16/3x^6 - 40x^4 + 100x^2- 274/3#.

Hopefully this helps!