Problem 67 - Calculus Early Transcendentals

Question

Find the solution of the following initial value problems. $$f^{\prime}(x)=2 x-3 ; f(0)=4$$

Step-by-Step Solution

Answer

Question: Determine the function f(x) that satisfies the initial value problem f'(x) = 2x - 3 and f(0) = 4. Answer: The function f(x) that satisfies the initial value problem is f(x) = x^2 - 3x + 4.

Step 1: Integrate the derivative

To find the function f(x), we need to integrate the given derivative f'(x). Integrate the function with respect to x: $$\int f'(x) dx = \int (2x - 3) dx$$

Step 2: Calculate the integral

Applying the power rule and integrating the constants, we get: $$f(x) = \int (2x - 3) dx = x^2 - 3x + C$$ where C is the constant of integration.

Step 3: Use the initial condition to find the constant

# Now, we'll use the initial condition f(0) = 4 to find the constant C. Substitute x=0 and f(x)=4 in the equation and solve for C: $$4 = (0)^2 - 3(0) + C \implies C = 4$$

Step 4: Write the particular solution

# Now that we have found the constant C, we can write the particular solution for the given initial value problem: $$f(x) = x^2 - 3x + 4$$ So, the solution to the initial value problem is: $$f(x) = x^2 - 3x + 4$$