notes / updates | ap calculus bc

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January 22, 2026 | Mean Value Theorem for Integrals (Demo)
https://www.geogebra.org/m/Jan0wXhp

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January 15, 2026 | Feedback on today's quiz
We know that \( f''(x) = 6x\). From the fact that the tangent line equation is: \( y = 5 - 3x \) at \(x=1 \),
we also know that \( f(1) = 2, f'(1) = -3 \)
Integrate: \( f'(x) = \int f''(x) dx = \int 6x dx = 3x^2 + C_1 \)
Plug in \( x=1, f'(1) = -3 \), to find \( C_1 = -6 \)
Integrate again: \( f(x) = \int f'(x) dx = \int (3x^2 -6) dx = x^3 -6x + C_2 \)
Plug in \(x=1, f(1) = 2 \), to find \( C_2 = 7 \)
Therefore, \( f(x) = x^3 - 6x + 7\).

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