Q.The least value of $f(x) = x^3 - 12x$, $x \in [0, 3]$ is (A) $-16$ (B) $-9$ (C) $0$ (D) $16$

When we need to find the maximum or minimum value of a function over a specific interval, especially a closed one, we are looking for its global extrema. For a continuous function on a closed interval, these global extrema are guaranteed to exist. The key insight from calculus is that these extreme values can only occur at two types of points:

1. Critical points: These are points where the derivative of the function is zero or undefined. At such points, the function might have a local maximum or a local minimum.
2. Endpoints of the interval: Even if the function is increasing or decreasing throughout the interval, its maximum or minimum value could be at one of the boundaries.

Therefore, our strategy is to find all potential candidates for the minimum (critical points within the interval and the endpoints) and then compare the function's value at each of these candidates. The smallest value will be the least value of the function on the given interval.

Let's apply this method to $f(x) = x^3 - 12x$ on the interval $x \in [0, 3]$.

1. Find the derivative of the function.

   To locate critical points, we first need the first derivative of $f(x)$.

   $f(x) = x^3 - 12x$

   $f'(x) = \frac{d}{dx}(x^3 - 12x) = 3x^2 - 12$

2. Find the critical points.

   Critical points occur where $f'(x) = 0$ or where $f'(x)$ is undefined. Since $f'(x) = 3x^2 - 12$ is a polynomial, it is defined for all real $x$. So, we only need to set $f'(x) = 0$:

   $3x^2 - 12 = 0$

   $3x^2 = 12$

   $x^2 = 4$

   Solving for $x$, we get $x = \pm 2$.

3. Identify critical points within the given interval.

   The given interval is $[0, 3]$. We must check which of our critical points fall within this range.

   • $x = 2$ is in $[0, 3]$. This is a candidate for the minimum.
   • $x = -2$ is not in $[0, 3]$. We discard this critical point for this problem, as it's outside our domain of interest. …

When we need to find the absolute maximum (or minimum) value of a continuous function over a closed interval, we rely on a fundamental concept from calculus called the Extreme Value Theorem. This theorem guarantees that such a maximum and minimum must exist.

The intuition behind finding these extreme values is that they can occur in one of two places:

1. At a "peak" or "valley" within the interval: These are points where the function changes from increasing to decreasing (local maximum) or decreasing to increasing (local minimum). At such points, if the function is differentiable, its derivative will be zero. These are called critical points.
2. At the boundaries of the interval: Even if the function is steadily increasing or decreasing throughout the interval, its highest or lowest value might simply be at one of the endpoints.

Therefore, our strategy is to check all these potential locations: the critical points that fall within our interval, and the two endpoints of the interval. We then compare the function values at all these points to find the absolute maximum.

Here's how we apply this to $f(x) = x^3 - 3x + 2$ on the interval $[0, 2]$:

1. Find the derivative of the function. The derivative $f'(x)$ tells us about the slope of the tangent line to the function at any point $x$. Critical points occur where the tangent line is horizontal, meaning $f'(x) = 0$.

2. Find the critical points by setting the derivative to zero. We solve $f'(x) = 0$ to find the $x$-values where the function might have a local maximum or minimum.

This is a difference of squares, which factors as $(x-1)(x+1)=0$.
So, the critical points are $x = 1$ and $x = -1$.

3. Identify which critical points lie within the given interval.

The given interval is $[0, 2]$. We must only consider critical points that are inside or on the boundary of this interval.

* $x = 1$ is in $[0, 2]$. This is a relevant critical point. …