The function f(x) = 2x³ – 3x² – 12x + 4 has

The correct option is "one maxima and one minima".

To find the local maxima and minima of the function, we use the first and second derivative tests. Let the given function be:
$$f(x)=2x^3-3x^2-12x+4$$

Step 1: Find the first derivative, $f^{\prime }\left(x\right)$.
Differentiating $f\left(x\right)$ with respect to $x$:
$$f^{\prime }(x)=\frac{d}{dx}(2x^3-3x^2-12x+4)$$
Using the power rule $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$:
$$f^{\prime }(x)=6x^2-6x-12$$

Step 2: Find the critical points by setting $f^{\prime }\left(x\right)=0$.
$$6x^2-6x-12=0$$
Divide the entire equation by 6:
$$x^2-x-2=0$$
Factor the quadratic equation:
$$(x-2)(x+1)=0$$
Thus, the critical points are $x=2$ and $x=-1$.

Step 3: Use the second derivative test to determine the nature of these points.
Find the second derivative, $f^{\prime \prime }\left(x\right)$:
$$f^{\prime \prime }(x)=\frac{d}{dx}(6x^2-6x-12)=12x-6$$

Now, we evaluate the second derivative at each critical point:
1. At $x=2$:
$$f^{\prime \prime }(2)=12(2)-6=24-6=18$$
Since $f^{\prime \prime }\left(2\right)>0$, the function has a local minimum at $x=2$.

2. At $x=-1$:
$$f^{\prime \prime }(-1)=12(-1)-6=-12-6=-18$$
Since $f^{\prime \prime }(-1)<0$, the function has a local maximum at $x=-1$.

Therefore, the function has exactly one maxima and one minima.