Show that for any constant k , the function f ( x ) = x - k has minimum value at x = k , though f ' ( x ) does not exist.

Given that f ( x ) = x - k where k is a constant Modulus function takes every real number to a non-negative real number. So, g ( y ) = y has the minimum value when y = 0 Thus, f ( x ) = x - k is minimum when x - k = 0 i.e., x = k We can rewrite the given function as f ( x ) = { - ( x - k ) x - k < 0 x - k x - k ≥ 0 i.e, f ( x ) = { k - x x < k x - k x ≥ k Now, f ' ( x ) for x < k is - 1 and f ' ( x ) for x ≥ k is 1 Right-hand derivative and left-hand derivative at x = k is not equal. Thus, the derivative of f ( x ) does not exist at x = k . So for any constant k , the function f ( x ) = x - k has minimum value at x = k , though f ' ( x ) does not exist.

If an odd function g ( x ) has a local minimum value at x = c where c ≠ 0 , can anything be said about the values of g at x = - c ? Give reasons for your answer.

It is given that g ( x ) is an odd function which has a local minimum value at x = c . Since, g ( x ) is an odd function, we have g ( - x ) = - g ( x ) . . . ( 1 ) g ( x ) has local minimum at x = c ⇒ There exists an open interval ( a , b ) containing c such that g ( x ) ≥ g ( c ) for all x ∈ ( a , b ) ⇒ - g ( x ) ≤ - g ( c ) ⇒ g ( - x ) ≤ g ( - c ) ...(from ( 1 ) ) As x ∈ ( a , b ) , we have - x ∈ ( - b , - a ) So, we have g ( - x ) ≤ g ( - c ) on the open interval ( - b , - a ) Hence, g ( x ) has the local maximum at x = - c

E M B I B E

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 6 - Application of Derivatives>Exercise

Embibe Experts Solutions for Chapter: Application of Derivatives, Exercise 1: Exercise

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Application of Derivatives, Exercise 1: Exercise

Attempt the free practice questions on Chapter 6: Application of Derivatives, Exercise 1: Exercise with hints and solutions to strengthen your understanding. Mathematics Crash Course (Based on Revised Syllabus-2023) solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Application of Derivatives, Exercise 1: Exercise with Hints & Solutions

EASY

12th Goa Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 6 - Application of Derivatives>Exercise>Q 4

Show that for any constant $k$ , the function $x < k$ has minimum value at $x = k$ , though $f^{'} (x)$ does not exist.

EASY

12th Goa Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 6 - Application of Derivatives>Exercise>Q 5

Find the values of $a$ and $b$ so that $f (x) = 4 x^{3} + a x^{2} + b x + 2$ may have an extreme value at $(2, 2)$ .

EASY

12th Goa Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 6 - Application of Derivatives>Exercise>Q 6

Show that the maximum value of the function $f (x) = 4 x - x^{2} - 1$ is $3$ .

MEDIUM

12th Goa Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 6 - Application of Derivatives>Exercise>Q 7

Show that the function $f (x) = x^{3} - 3 x^{2} + 9 x + 5$ possesses no maximum or minimum value.

EASY

12th Goa Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 6 - Application of Derivatives>Exercise>Q 8

Examine whether the function $f (x) = {(x - 2)}^{2} + 4 x$ has any maximum or minimum value at any point in its domain of definition.

EASY

12th Goa Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 6 - Application of Derivatives>Exercise>Q 9

If an odd function $g (x)$ has a local minimum value at $x = c$ where $c \neq 0$ , can anything be said about the values of $g$ at $x = - c$ $?$ Give reasons for your answer.