The maximum value of f(x) = 2x3 – 9x2 + 12x – 3 in the interval 0 ≤ x ≤ 3 is _______
Concept:
The point of maxima or minima is obtained by solving for the derivative of the function and equating to zero.
Then, to check if the point is a point of maxima ‘or’ minima we check the second derivative at that point.
This is explained with the help of the following graph:
If \(\frac{{{d^2}f}}{{d{x^2}}} < 0\); the point will be a point of maxima
If \(\frac{{{d^2}f}}{{d{x^2}}} > 0\), the point will be a point of minima.
Analysis:
Since we are given an interval, we will obtain a local maximum or minimum in the interval.
Solving for the first derivative and equating it to zero, we get:
f'(x) = 6x2 – 18x + 12 = 0
x2 – 3x + 2 = 0
(x - 1)(x - 2) = 0
X = 1 and x = 2
∴ x = 1 and x = 2 are possible local minimum and maximum in the interval 0 and 3.
Checking for the double derivative at these points, we’ll get:
f’’(x) = 12x – 18
f’’(1) = 12 – 18 = -6
f’’(1) < 0
∴ x = 1 is a local maximum
Similarly,
f’’(2) = 12 × 2 – 18 = 6
Since, f’’(2) > 0, x = 2 is a local minimum.
Since we are to find maximum and minimum value in the interval 0 ≤ x ≤ 3, we will check the border conditions for local maximum as well:
x | f(x) = 2x3 – 9x2 + 12x – 3 | 0 ≤ x ≤ 3 |
0 1 2 3 | f(x) = -3 f(x) = 2 f(x) = 1 f(x) = 6 | Local maximum Local minimum Maximum value |
∴ The maximum value of f(x) in the interval 0 ≤ x ≤ 3 will be 6.
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 minimum value of the function f(x) = x3 – 3x2 – 24 x + 100 in the interval [-3, 3] is
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 )
- For 0 ≤ t < ∞, the maximum value of the function f(t) = e-t – 2e-2t occurs at:
