Find $f'(x)$ and $f''(x)$: \[ f'(x) = 3x^2 - 12x + 12, \quad f''(x) = 6x - 12. \] At $x = 2$: \[ f'(2) = 0, \quad f''(2) = 0. \] Check $f'''(x)$: \[ f'''(x) = 6 \implies f'''(2) = 6 \neq 0. \] Thus, $x = 2$ is a point of inflection.
Area of region enclosed by curve y=x3 and its tangent at (–1,–1)
The minimum of \(f(x)=\sqrt{(10-x^2)}\) in the interval \([-3,2]\) is