Question

Let f be a differentiable function such that f(1) = 2 and f'(x) = x^2 - 2x + 3. Find:

a) f(0)

b) f(2)

c) ∫ (x^2 - 2x + 3) dx

d) f''(x)

Answer 1

To find f(0), we need to integrate from 1 to 0: ∫ (x^2 - 2x + 3) dx. Plugging in the limits, we get f(0) = ∫ (x^2 - 2x + 3) dx = (-1/3)x^3 + x^2 + 3x + C. To find f(2), we can use the same integral but with limits from 1 to 2: ∫ (x^2 - 2x + 3) dx. Plugging in the limits, we get f(2) = ∫ (x^2 - 2x + 3) dx = (-1/3)x^3 + x^2 + 3x + C. This integral gives us the original function f(x). Simplifying the integral, we get ∫ (x^2 - 2x + 3) dx = (1/3)x^3 - x^2 + 3x + C. To find the second derivative of f(x), we can differentiate the given derivative function: f''(x) = 2x - 2.

Step-by-step explanation:

To find the answers, we need to integrate the given derivative function:

a) f(0): To find f(0), we need to integrate from 1 to 0: ∫ (x^2 - 2x + 3) dx. Plugging in the limits, we get f(0) = ∫ (x^2 - 2x + 3) dx = (-1/3)x^3 + x^2 + 3x + C. Plugging in the value of f(1) = 2, we can solve for C and find f(0).

b) f(2): To find f(2), we can use the same integral but with limits from 1 to 2: ∫ (x^2 - 2x + 3) dx. Plugging in the limits, we get f(2) = ∫ (x^2 - 2x + 3) dx = (-1/3)x^3 + x^2 + 3x + C. Plugging in the value of f(1) = 2, we can solve for C and find f(2).

c) ∫ (x^2 - 2x + 3) dx: This integral gives us the original function f(x). Simplifying the integral, we get ∫ (x^2 - 2x + 3) dx = (1/3)x^3 - x^2 + 3x + C.

d) f''(x): To find the second derivative of f(x), we can differentiate the given derivative function: f''(x) = 2x - 2.