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