Question
Let f(x) be a real -valued function such that f'(x0) = 0 for some x0 ∈ (0, 1), and f"(x0) > 0 for all x ∈ (0, 1). Then f(x) has
Options :
No local minimum in (0, 1)
One local maximum in (0, 1)
Exactly one local minimum in (0, 1)
Two distinct local minimum in (0, 1)
Answer :
Exactly one local minimum in (0, 1)
Solution :
Concept:
Consider a function y = f(x) on a defined interval of x. The function attains extreme values (the value can be maximum or minimum or both).
For maxima:
Local maxima: A point is the local maxima of a function if there is some other point where the maximum value is greater than the local maxima but that point doesn’t exist nearby local maxima.
Global maxima: It is the point where there is no other point has in the domain for which function has more value than global maxima.
For minima:
Local minima: A point is the local minima of a function if there is some other point where the minimum value is less than the local minima but that point doesn’t exist nearby local minima.
Global minima: It is the point where there is no other point has in the domain for which function has less value than global minima.
Stationary Points:
Points where the derivative of the function is zero i.e., f’(x) = 0.
The points can be: Inflection point ,Local maxima, Local minima
Second derivative test: Let the function has a stationary point x = a
if x=-1 and x=2 are extreme points
If then x = a, is a point of minima.
Application:
Given f(x) is a real -valued function such that f'(x0) = 0 for some x0 ∈ (0, 1) Also given f"(x) > 0 for all x ∈ (0, 1) So,
Then f(x) has exactly one local minimum in (0, 1), called the point of minima.
Copyright © 2025 Test Academy All Rights Reserved