Criteria For Relative Extrema
Criteria For Relative Extrema Assignment Help | Criteria For Relative Extrema Homework Help
Criteria For Relative Extrema
Let f ‘(x0) = f ‘’ (x0) = ... = f (n-1) (x0) = 0, but f(n) (x0) ≠ 0. Then
1. If n is odd, f has neither a relative maximum nor a relative minimum at x = x0.
2. If n is even, f has a relative maximum if f (n) (x0) < 0, and has a relative minimum if f (n) (x0) > 0.
This method is illustrated in the following examples.
Example. Use the second derivative test to find the relative maxima and minima of f (x) = 2x3 – 15x2 + 36x + 18.
Solution. Differentiating the given function with respect to x, we obtain
f ‘ (x) = 6x2 – 30x + 36 = 6 (x-2) (x-3)
Now f ‘ (x) = 0 when x = 2 and x = 3
Also f “ (x) = 12x – 30
f “ (2) = 12 (2) – 30 = - 6 < 0
hence there is a relative maximum at x = 2.
Further f “ (3) = 12 (3) – 30 = 6 > 0,
hence there is a relative minimum at x = 3.
For more help in Criteria For Relative Extrema click the button below to submit your homework assignment
1. If n is odd, f has neither a relative maximum nor a relative minimum at x = x0.
2. If n is even, f has a relative maximum if f (n) (x0) < 0, and has a relative minimum if f (n) (x0) > 0.
This method is illustrated in the following examples.
Example. Use the second derivative test to find the relative maxima and minima of f (x) = 2x3 – 15x2 + 36x + 18.
Solution. Differentiating the given function with respect to x, we obtain
f ‘ (x) = 6x2 – 30x + 36 = 6 (x-2) (x-3)
Now f ‘ (x) = 0 when x = 2 and x = 3
Also f “ (x) = 12x – 30
f “ (2) = 12 (2) – 30 = - 6 < 0
hence there is a relative maximum at x = 2.
Further f “ (3) = 12 (3) – 30 = 6 > 0,
hence there is a relative minimum at x = 3.
For more help in Criteria For Relative Extrema click the button below to submit your homework assignment