The function f(x) = 2×3 – 3×2 – 36x + 2 has its maxima at A. x = -2 only B. x = 0 only C. x = 3 only D. both x = -2 and x = 3

x = -2 only
x = 0 only
x = 3 only
both x = -2 and x = 3

The function $f(x) = 2x^3 – 3x^2 – 36x + 2$ has its maxima at $x = -2$ and $x = 3$.

To find the maxima and minima of a function, we can use the second derivative test. The second derivative of $f(x)$ is $f”(x) = 12(x + 2)(x – 3)$.

The second derivative is equal to zero at $x = -2$ and $x = 3$. Since the second derivative is a polynomial, it is defined for all real numbers. Therefore, the only possible critical points of $f(x)$ are $x = -2$ and $x = 3$.

To determine whether a critical point is a maximum or minimum, we can evaluate the sign of the second derivative at that point. If the second derivative is positive at a critical point, then the function has a minimum at that point. If the second derivative is negative at a critical point, then the function has a maximum at that point.

In this case, the second derivative is positive at $x = -2$ and negative at $x = 3$. Therefore, $f(x)$ has a minimum at $x = -2$ and a maximum at $x = 3$.