Show that the least value of the function `f(x)=2x^3-3x^2-12x+1` on `[-2,2.5]` has one maxima and one minima.

Explanation

To show that the least value of the function \( f(x) = 2x^3 - 3x^2 - 12x + 1 \) on the interval \([-2, 2.5]\) has one maximum and one minimum, we will follow these steps:

### Step 1: Find the derivative of the function
To find the critical points, we first differentiate the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 12x + 1) \]
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