Derivatives Opposite

Question

Find f.

f ''(x) = −2 + 30x − 12x²,    f(0) = 2,    f '(0) = 14

f(x) = ?

Answer 1

f"(x) = -2 + 30x - 12x²

Integrate to get f'(x) = -2x + 15x² - 4x³ + C₁

Since f'(0) = 14, we have C₁ = 14. So, f'(x) = -2x + 15x² - 4x³ + 14

Integrate again to get f(x) = -x⁴ + 5x³ - x² + 14x + C₂

Since f(0) = 2, C₂ = 2. So, f(x) = -x⁴ + 5x³ - x² + 14x + 2

Answer 2

"Reverse, reverse!"

Let's start with f"(x). If you're the physics type, you'll recognize this function as an acceleration function. We need to get to the level above it (velocity) and to the level above that (position). We have to satisfy a couple of conditions along the way, though. (Think of an initial value problem as a kind of wish-list that a family of antiderivatives may have for you. Ready, Santa?)

f ''(x) = −2 + 30x − 12x²,    f(0) = 2,    f '(0) = 14

f(x) = ?

Start with f"(x) = -2 + 30x - 12x². Integration (that is, anti-differentiation) is a linear operator, so integrate all the way across. The antiderivative of -2 is -2x. This is because the antiderivative of a constant is that same constant multiplied by the independent variable (here, x). The antiderivative of 30x is 15x². This is because of the power rule of integration. The antiderivative of -12x² is -4x³, and this is also by the power rule for integration. You can verify that these are correct is taking the derivatives of the antiderivatives. Do you get back the terms of f"(x)? Good.

So the antiderivative of f"(x) is f'(x) = -2x + 15x² - 4x³ + C . This is a velocity function, and it is an indefinite integral since we're not capturing velocity between two specific, pre-defined times. It is very general, so we need to account for any constant that could exist in the antiderivative. We also need that C to satisfy the conditions of the original problem. So, let's satisfy our first initial condition. We need f'(0) = 14. Replace f'(x) with 14 in the function we just found, and plug in 0 wherever you see "x" in the right-hand side. You'll get ( 14 ) = -2(0) + 15(0)² - 4(0)³ + C. It follows that C = 14.

The updated velocity model is f'(x) = -2x + 15x² - 4x³ + 14. Now we integrate this function to get back to a position function. Using constant and power rules for integration similarly as above, we get:

f(x) = -x² + 5x³ - x⁴ + 14x + C. And similarly as above, we are not trying to find displacement between two specific, pre-defined times. This is an indefinite integral, which is why I have a C. This will be used to satisfy the next initial value condition. We need f(0) = 2. Replace the f(x) with 2, and plug in 0 wherever you see "x" in the right-hand side. You'll get ( 2 ) = -(0)² + 5(0)³ - (0)⁴ + 14(0) + C. It follows that C = 2.

The updated displacement model is f(x) = -x² + 5x³ - x⁴ + 14x + 2.

Let's clean it up a bit:

f(x) = -x⁴ + 5x³ - x² + 14x + 2.

What happens when I plug in 0 straight into f(x)? All terms except the constant 2 cancel out. That is, f(x) = 2. Awesome! That's a condition satisfied.

If I take the derivative of f(x) and plug in 0 when I get there, I get:

-4x³ + 15x² - 2x + 14 ----(plug in 0)---> 14. Awesome! That's a condition satisfied.

If I take the second derivative of f(x), I get:

-12x² + 30x - 2. Just like in the original problem.

I've satisfied all that the problem has demanded of me. I leave this function as the final answer:

f(x) = -x⁴ + 5x³ - x² + 14x + 2. Please let me know if I need to clarify any of the integration or differentiation procedures.