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f(x) is a real valued function then
if f''(a)>0 then function is said to be decresing at x=a
if f''(x)<0 then function is said to be increasing at x=a
from where it comes?
Dear vikash,
Let f be continuous on [a, b] and differentiable on the open interval (a, b). Then
(a) f is increasing on [a, b] if f '(x) > 0 for each x (a, b)
(b) f is decreasing on [a, b] if f '(x) < 0 for each x (a, b)
This theorem can be proved by using Mean Value Theorem. We shall prove the theorem after learning Mean Value Theorem.
This theorem is applied in various problems to check whether a function is increasing or decreasing.
(1) Let the given function be f (x) on the real number line R.
(2) Differentiate the function f(x) with respect to x and equate it to zero i.e., put f '(x) = 0. Solve for x. These values of x which satisfy f '(x) = 0 are called Critical values of the function
(3) Arrange these Critical values in ascending order and partition the domain of f (x) into various intervals, using the Critical values.
(4) Check the sign of f '(x) in each open intervals.
(5) If f '(x) > 0 in a particular interval, then the function is increasing in that particular interval.
If f '(x) < 0 in a particular interval, then the function is decreasing in that particular interval.
We are all IITians and here to help you in your IIT JEE preparation. All the best. If you like this answer please approve it.... win exciting gifts by answering the questions on Discussion Forum Sagar Singh B.Tech IIT Delhi
We are all IITians and here to help you in your IIT JEE preparation. All the best.
If you like this answer please approve it....
win exciting gifts by answering the questions on Discussion Forum
Sagar Singh
B.Tech IIT Delhi
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