Question

The function $f(x)=2x^3-3x^2-12x+4$, has

• A. 2 points of local maxima
• B. 2 points of local minimum
• C. One maxima and one minima
• D. No maxima nor minima

Answer 1

Answer: (C)

One maxima and one minima

Answer 2

To find the local maxima and minima of the function $f(x)=2x^3-3x^2-12x+4$, we follow these steps:

1. Find the first derivative of the function, $f'(x)$: $f(x) = 2x^3-3x^2-12x+4$ $f'(x) = \frac{d}{dx}(2x^3) - \frac{d}{dx}(3x^2) - \frac{d}{dx}(12x) + \frac{d}{dx}(4)$ $f'(x) = 6x^2 - 6x - 12$

2. Find the critical points by setting $f'(x) = 0$: $6x^2 - 6x - 12 = 0$ Divide the entire equation by 6: $x^2 - x - 2 = 0$ Factor the quadratic equation: $(x-2)(x+1) = 0$ This gives us two critical points: $x = 2$ and $x = -1$.

3. Use the second derivative test to classify the critical points: First, find the second derivative, $f''(x)$: $f'(x) = 6x^2 - 6x - 12$ $f''(x) = \frac{d}{dx}(6x^2) - \frac{d}{dx}(6x) - \frac{d}{dx}(12)$ $f''(x) = 12x - 6$

   Now, evaluate $f''(x)$ at each critical point:

   • For $x = -1$: $f''(-1) = 12(-1) - 6 = -12 - 6 = -18$ Since $f''(-1) < 0$, there is a local maximum at $x = -1$.

   • For $x = 2$: $f''(2) = 12(2) - 6 = 24 - 6 = 18$ Since $f''(2) > 0$, there is a local minimum at $x = 2$.

Conclusion:

The function $f(x)$ has one local maximum (at $x=-1$) and one local minimum (at $x=2$).