Mth101 lec 23 to 45 by kinza bilal-1

Lecture 23: Maximum & Minimum Values of Function

Absolute Maximum and Minimum

  • Absolute Maximum: If ( f(x_0) \geq f(x) ) for all ( x ) in the domain of ( f ), then ( f(x_0) ) is called the absolute maximum value or simply the maximum value.

  • Absolute Minimum: If ( f(x) \leq f(x_0) ) for all ( x ) in the domain of ( f ), then ( f(x_0) ) is called the absolute minimum value or simply the minimum value.

  • Remark: Extremum refers to the maximum or minimum value over the entire domain of ( f ).

Example Calculation

  • Find the maximum and minimum values of ( f(x) = 2x + 1 ) over the interval [0, 3):

    • Evaluate at boundary points:

      • ( x = 0 ): ( f(0) = 2(0) + 1 = 1 )

      • ( x = 2 ): ( f(2) = 2(2) + 1 = 5 )

    • Absolute minimum value is ( 1 ) at ( x = 0 ).

    • There's no maximum value since the function approaches but does not reach its peak in this interval.

Lecture 24: Extreme Value Theorem (EVT)

Theorem Overview

  • If a function is continuous on a closed interval ( [a, b] ), then it attains both a maximum and a minimum value on ( [a, b] ).

  • Example: In prior example, ( f(x) = 2x + 1 ) is continuous over ( [0, 3) ) but does not meet EVT conditions as it does not include endpoint 3.

Critical Points

  • A function ( f ) has an extremum on an open interval ( (a, b) ) if it occurs at a critical point, defined by derivatives equating to zero.

Lecture 25: Steps to Find Extremum

Steps for Finding Maximum and Minimum Values

  1. Find Critical Points: Determine ( f'(x) = 0 ) within ( (a, b) ).

  2. Evaluate Function: Calculate ( f ) at all critical points and endpoints.

  3. Determine Maximum: Identify the largest value from calculated points as maximum value.

  4. Determine Minimum: Identify the smallest value from calculated points as minimum value.

Example Calculation

  • For ( f(x) = 2x^3 - 15x + 36 ) on interval ( [1,5] ):

    • Step 1: Calculate the derivative: ( f'(x) = 6x^2 - 15 ), set to 0 to find critical points.

    • Step 2: Evaluate ( f ) at critical points ( x = 2 ) and ( x = 3 ).

      • ( f(2) = 28 )

      • ( f(3) = 24 )

      • Endpoints: ( f(1) = 23, f(5) = 54 )

    • Maximum at ( x = 5 ): ( f(5) = 54 ), minimum at ( x = 1 ): ( f(1) = 23 ).

Lecture 26: Applied Minimum & Maximum Problems

Example Problem: Rectangle Area Maximization

  • Problem: Find dimensions of a rectangle with a perimeter of 100 ft to maximize the area.

    • Let ( x ) be the length and ( y ) be the width. Area: ( A = xy ) and Perimeter: ( 2x + 2y = 100 ).

    • Substitute: ( y = 50 - x )

    • Area Formula: ( A = x(50 - x) = 50x - x^2 )

    • Differentiate: ( A' = 50 - 2x ), set equal to zero to find critical points.

      • Critical point at ( x = 25 ) yields maximum area: 625 sq ft.

Lecture 27: Sigma Notation

  • Sigma Notation: A compact way to depict sums over a series of terms. Example:

    • The notation ( \sum_{k=1}^{n} k^2 ) denotes the sum of squares of the first n natural numbers.

  • Upper & Lower Limits: The limits in sigma notation define the range of summation, such as starting at 1 and ending at n.

Changing Index of Summation

  • To change the index in a sigma expression, create suitable substitutions to keep the sum equivalent.

Lecture 28: Integration Area Problems

Area Under a Curve

  • The area can be calculated via limits of sums, turning to integral calculus.

  • Example: Find the area under the line ( y = x ) over the interval [1, 2] by calculating limits through evaluation.

    • Use a Riemann sum or the definite integral.

Lecture 29: Definite Integral

Area using Definite Integrals

  • Definition: For a function ( f ) continuous on ( [a, b] ), the area A under the curve can be represented by ( A = ext{lim}_{n o ext{∞}} ext{Riemann Sum} )

  • Explore properties like switching integration limits and the effect of constant factors.

Lecture 30: Fundamental Theorems of Calculus

First Fundamental Theorem

  • For a continuous function ( f ) on ( [a, b] ), if ( F(x) ) is an antiderivative, then:

    • ( \int_{a}^{b} f(x)dx = F(b) - F(a) )

Example of Evaluation

  • Calculate ( \int_{2}^{4} 2x dx ) using the first fundamental theorem.

Lecture 31: Evaluating Definite Integrals

Techniques & Methods

  • Use substitution methods to evaluate integrals, adjusting limits accordingly and maintaining proper notation.

Lecture 32: The Second Fundamental Theorem

Importance

  • If the integrand is continuous, the derivative of the definite integral with variable limits gives back the original function.

Continuing Pages

  • Proceeding through lectures 33-45 covers topics like Volume Calculations, Series definitions, Rational Tests, and Advanced Calculus concepts such as Taylor & Maclaurin series.