John W. answered • 07/22/20
Bachelor's Degree in Chemical Engineering
First, let's integrate the second derivative twice to find a general form for f. Integrating once, we get f'(x) = 6x + 3x2 + 8x3 + C1, and integrating again we get f(x) = 3x2 + x3 + 2x4 + C1x + C2. Now that we have the general form of f(x), we need to figure out the values of the constants C1 and C2, which we can do with the remaining information given in the problem. First, let's use the equation f(0) = 4. Plugging in 0 for x into our expression for f(x) gives 3(0)2 + (0)3 + 2(0)4 + C1(0) + C2 which evaluates to just C2. So, f(0) = C2 = 4. We can rewrite f(x) now as f(x) = 3x2 + x3 + 2x4 + C1x + 4. Now, let's find C1 using f(1) = 15. Plugging in 1 for x into our expression for f(x) gives 3(1)2 + (1)3 + 2(1)4 + C1(1) + 4 which evaluates to C1 + 10. So, f(1) = 10 + C1 = 15, and so C1 = 5. Thus, our final answer is f(x) = 3x2 + x3 + 2x4 + 5x + 4. Typically we write polynomial functions with the highest exponents first and the lowest exponents last, so you may want to write it as f(x) = 2x4 + x3 + 3x2 + 5x + 4. You can also check the expression by differentiating twice to find the second derivative and comparing it to the problem statement, and you can plug in x = 0 and x = 1 to check that the function gives 4 and 15 respectively.
