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

The function is

I would recommend expanding and then integrating.

Write in Leibniz Notation:

We now have a separable differential equation.

Solve for C now:

Hopefully this helps!

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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

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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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