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.

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All the best.

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