Question:

Given the second derivative of the curve $y=f(x)$ as $\dfrac{d^2y}{dx^2}=6x-4$, and that it has a local minimum value $5$ at $x=1$. Step 1: Integrate the second derivative to find the first derivative. $$f''(x) = 6x-4$$ $$f'(x) = \int (6x-4) dx = 3x^2 - 4x + C_1$$ Step 2: Use the condition for a local minimum to determine $C_1$. Since there is a local minimum at $x=1$, we know that $f'(1)=0$. $$3(1)^2 - 4(1) + C_1 = 0$$ $$3 - 4 + C_1 = 0$$ $$-1 + C_1 = 0$$ $$C_1 = 1$$ Thus, the first derivative is $f'(x) = 3x^2 - 4x + 1$. Step 3: Integrate the first derivative to find the function $f(x)$. $$f(x) = \int (3x^2 - 4x + 1) dx = x^3 - 2x^2 + x + C_2$$ Step 4: Use the local minimum value to determine $C_2$. The local minimum value is $5$ at $x=1$, which means $f(1)=5$. $$(1)^3 - 2(1)^2 + (1) + C_2 = 5$$ $$1 - 2 + 1 + C_2 = 5$$ $$0 + C_2 = 5$$ $$C_2 = 5$$ Thus, the function is $f(x) = x^3 - 2x^2 + x + 5$. Step 5: Calculate $f(0)$. Substitute $x=0$ into the function $f(x)$: $$f(0) = (0)^3 - 2(0)^2 + (0) + 5$$ $$f(0) = 0 - 0 + 0 + 5$$ $$f(0) = 5$$