find the equation of the function that has 𝑓′(𝑥)=6𝑥3+𝑥+7𝑓′(𝑥)=6𝑥3+𝑥+7 and passes through the point (0,4)

The given derivative function is f'(x) = 6x^3 + x + 7.

To find the original function f(x), we need to integrate the derivative function.

∫f'(x) dx = ∫(6x^3 + x + 7) dx = (6/4)x^4 + (1/2)x^2 + 7x + C.

So, f(x) = (3/2)x^4 + (1/2)x^2 + 7x + C.

We know that the function passes through the point (0,4), which means when x=0, f(x)=4. We can use this information to find the constant C.

4 = (3/2)(0)^4 + (1/2)(0)^2 + 7*(0) + C 4 = C.

So, the equation of the function is f(x) = (3/2)x^4 + (1/2)x^2 + 7x + 4.