Notes
Example 1: Find all the points of local maxima and local minima of the function f given by f(x) = 2x^{3}  6x^{2} + 6x + 5
Solution: Given that f(x) = 2x^{3}  6x^{2} + 6x + 5
f ‘(x) = 6x^{2}  12x + 6 = 6(x  1)^{2}
f ‘(x) = 0 at x = 1
Observe that f ‘(x) > 0 for all x R and in particular f ‘(x) > 0, for values close to 1 and to the left and right of 1.
Hence x = 1 is a point of inflection.
Let f be a function defined on an interval I and c I. Let f be twice differentiable at c.
Then
Example 2: Find local maximum and local minimum values of the function f given by
f(x) = 3x^{4} + 4x^{3}  12x^{2} + 12
Solution: f(x) = 3x^{4} + 4x^{3}  12x^{2} + 12
f ‘(x) = 12x^{3} + 12x^{2} – 24x = 12x(x – 1)(x + 2)
f ‘(x) = 0 at
x = 0, x = 1 and x = –2
f ‘'(x) = 36x^{2} + 24x – 24
f ‘'(0) = –24 < 0
f ‘'(1) = 36 > 0
f ‘'(–2) = 144 – 48 – 24 = 72 > 0
x = 0 is a point of local maxima and local maximum value of f(x) at x = 0 is f(0) = 12 while x = 1 and x = –2 are local minimum points.
Local minimum values are f(1) = 7 and f(–2) = 20
Step 2: Take the end points of the interval.
Step 3: At all these points (listed in step 1 and step 2) calculate the values of f(x).
Step 4: Identity the maximum and minimum values of f(x) out of the values will be the absolute maximum (greatest) value of f(x) and the minimum (least) value of f(x).
Example 3: Find the absolute maximum and minimum values of a function f(x) given by f(x) = 2x^{3} – 15x^{2} + 36x + 1 in the interval [1, 5]
Solution: f '(x) = 6x^{2} – 30x + 36 = 6(x – 3)(x – 2)
f '(x) = 0 gives x = 2, x = 3
We shall evaluate the f(x) at x = 2 and x = 3 and at the end points.
f (1) = 24
f(2) = 29
f(3) = 28
f(5) = 56
So absolute maximum value is 56 at x = 5
Absolute minimum value is 24 at x = 1
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