The minimum value of the function f(x) = x3 – 3x2 – 24 x + 100 in the interval [-3, 3] is
28
Finding Minimum Function Value in Interval
We are asked to find the minimum value of the function $f(x) = x^3 - 3x^2 - 24x + 100$ within the closed interval $[-3, 3]$. To do this, we will use calculus methods involving derivatives and evaluate the function at critical points and interval endpoints.
Step 1: Calculate the Derivative
First, we find the derivative of the function $f(x)$ with respect to $x$ to determine the function's rate of change.
The derivative, denoted as $f'(x)$, is:
$$ f'(x) = \frac{d}{dx}(x^3 - 3x^2 - 24x + 100) $$
$$ f'(x) = 3x^2 - 6x - 24 $$
Step 2: Find Critical Points
Critical points occur where the derivative is zero or undefined. Since $f'(x)$ is a polynomial, it is defined for all $x$. We set $f'(x) = 0$ to find the critical points.
$$ 3x^2 - 6x - 24 = 0 $$
To simplify, we can divide the entire equation by 3:
$$ x^2 - 2x - 8 = 0 $$
Now, we factor the quadratic equation:
$$ (x - 4)(x + 2) = 0 $$
This gives us two critical points: $x = 4$ and $x = -2$.
Step 3: Identify Critical Points within the Interval
The given interval is $[-3, 3]$. We need to check which of the critical points fall within this interval.
- $x = 4$: This value is outside the interval $[-3, 3]$.
- $x = -2$: This value is inside the interval $[-3, 3]$.
Therefore, the only critical point we need to consider within the interval is $x = -2$.
Step 4: Evaluate the Function at Endpoints and Critical Point
According to the Extreme Value Theorem, the minimum (and maximum) value of a continuous function on a closed interval occurs either at the endpoints of the interval or at the critical points within the interval. We evaluate $f(x)$ at $x = -3$, $x = 3$ (the endpoints), and $x = -2$ (the critical point within the interval).
We can use a table to organize the calculations:
| x | Calculation of f(x) | f(x) Value |
| -3 | $f(-3) = (-3)^3 - 3(-3)^2 - 24(-3) + 100$ $f(-3) = -27 - 3(9) + 72 + 100$ $f(-3) = -27 - 27 + 72 + 100$ $f(-3) = -54 + 172$ |
118 |
| -2 | $f(-2) = (-2)^3 - 3(-2)^2 - 24(-2) + 100$ $f(-2) = -8 - 3(4) + 48 + 100$ $f(-2) = -8 - 12 + 48 + 100$ $f(-2) = -20 + 148$ |
128 |
| 3 | $f(3) = (3)^3 - 3(3)^2 - 24(3) + 100$ $f(3) = 27 - 3(9) - 72 + 100$ $f(3) = 27 - 27 - 72 + 100$ $f(3) = 0 - 72 + 100$ |
28 |
Step 5: Determine the Minimum Value
Now, we compare the values of $f(x)$ calculated at the relevant points:
- $f(-3) = 118$
- $f(-2) = 128$
- $f(3) = 28$
The smallest value among these is 28.
Therefore, the minimum value of the function $f(x) = x^3 - 3x^2 - 24x + 100$ in the interval $[-3, 3]$ is 28.
Important Questions from Maxima & Minima
Which of the following statements is false about convex minimization problem?
For what value of 'x' will the function y = x2 - 4x have the maximum or minimum value?
The function f(x) = 8 loge x - x2 + 3 attains its global minimum over the interval [1, e] at x = ________.
(Here loge x is the natural logarithm of x and e2 = 7.39 )
The maximum value of f(x) = 2x3 – 9x2 + 12x – 3 in the interval 0 ≤ x ≤ 3 is _______
- For 0 ≤ t < ∞, the maximum value of the function f(t) = e-t – 2e-2t occurs at:
