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

Question

Paisley Johnson · Accepted Answer

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!

Emma Aldridge · Answer

undefined To find an equation for the function f'(x) = 2x(4x^2-10)^2 that passes through the point (2,10), we can integrate f'(x) to obtain f(x).

First, integrate f'(x) with respect to x:
∫ f'(x) dx = ∫ 2x(4x^2-10)^2 dx

Next, simplify the integral:
f(x) = ∫ 2x(16x^4 - 80x^2 + 100) dx

Now, integrate each term separately:
f(x) = 2 ∫ (16x^5 - 80x^3 + 100x) dx

Integrating each term:
f(x) = 2 * (16/6)x^6 - (80/4)x^4 + (100/2)x^2 + C

Simplifying:
f(x) = (32/3)x^6 - 20x^4 + 50x^2 + C

To find the value of C, we use the fact that the graph passes through the point (2,10):
10 = (32/3)(2^6) - 20(2^4) + 50(2^2) + C

Solving for C:
C = 10 - (32/3)(64) + 20(16) - 50(4)

Simplifying:
C = 10 - 682.67 + 320 - 200

C ≈ -552.67

Therefore, the equation for the function f(x) = 2x(4x^2-10)^2 that passes through the point (2,10) is:
f(x) = (32/3)x^6 - 20x^4 + 50x^2 - 552.67