The smallest value of the function \[f(x) = 2x^3 − 9x^2 + 12x + 3\] in the range \(0 \leq x \leq 2\) is

\(1\),

\(3\),

\(5\),

\(7\).

The smallest value will occur either at a stationary point of \(f(x)\), or at one of the endpoints of the range, \(x = 0\) or \(x = 2\).

To find stationary points, we solve \(f'(x) = 6x^2 - 18x + 12 = 0\), or equivalently \(0 =x^2 - 3x + 2 = (x - 1)(x - 2)\).

So \(x = 1\) and \(x = 2\) are the stationary points of \(f(x)\), and are in the desired range. From our knowledge of the shape of cubics, or from looking at \(f''(x)\), we know that \(x = 1\) is a maximum and \(x = 2\) is a minimum.

Then the minimum value of \(f(x)\) is either at \(x = 2\) or \(x = 0\), so we can check the values. We have \[\begin{align*} f(2) &= 16 - 36 + 24 + 3 = 7 \\ f(0) &= 3. \end{align*}\]So the minimum value of \(f(x)\) for \(0 \leq x \leq 2\) is \(3\), and the answer is (b).