Final answer:
By integrating the second derivative and using the given conditions, the original function f(t) is determined, which doesn't exactly match any of the provided options, suggesting a possible typo or miscalculation.
Step-by-step explanation:
The student's question involves finding the original function f(t) given the second derivative f″(t) = 3t, and additional information about the function's value and first derivative at t=4: f(4) = 15 and f′(4) = 6. To find f(t), we need to integrate the second derivative twice and use the given conditions to solve for the constants.
First, integrate f″(t) = 3t to get the first derivative f′(t). The integral of 3t with respect to t is 3/2 t² + C. Then, use the condition f′(4) = 6 to find C:
f′(4) = (3/2) • 4² + C = 6
C = 6 - (3/2)• 4² = 6 - 24 = -18
So the first derivative is f′(t) = 3/2 t² - 18. Now, integrate f′(t) to obtain f(t):
f(t) = ∫ (3/2 t² - 18) dt = 1/2 t³ - 18t + C1
Then, use the condition f(4) = 15 to find C1:
f(4) = 1/2•4³ - 18•4 + C1 = 15
C1 = 15 - (1/2•64 - 18•4) = 15
Finally, f(t) is f(t) = 1/2 t³ - 18t + 15, which simplifies to f(t) = 1/2 t³ - 18t + 15. This function isn't explicitly listed in the options provided, suggesting there may have been a typo or calculation error either in this solution process or within the options given by the student.