f ( x ) = 2 x 2 - 8 x + 5 for x ∈ ℝ , 0 ⩽ x ⩽ k . For k = 2 . Find the range of f .

Given f ( x ) = 2 x 2 - 8 x + 5 Express function in the form a x + b 2 + c where a , b and c are constants To solve a x 2 + b x + c = 0 by completing the square: If a , the leading coefficient (the coefficient of the x 2 term), is not equal to 1 , take a as common factor ⇒ f x = 2 x 2 - 4 x + 5 2 Add and subtract the square of half the coefficient of the x -term, ⇒ f x = 2 x 2 - 4 x + 2 2 - 2 2 + 5 2 Factor the expression as a square of a binomial. ⇒ f x = 2 x - 2 2 - 2 2 + 5 2 ⇒ f x = 2 x - 2 2 - 3 2 ⇒ f x = 2 x - 2 2 - 3 , for x ∈ ℝ , 0 ≤ x ≤ 2 The part of the above expression is a square it will always be ≥ 0 . The smallest value it can be is 0 . This occurs when x = 2 . Thus, minimum value of f x occurs when x = 2 Therefore, when x = 2 then f ( x ) = 2 2 - 2 2 - 3 = - 3 is minimum When x = 1 then f ( x ) = 2 1 - 2 2 - 3 = - 1 When x = 0 then f ( x ) = 2 0 - 2 2 - 3 = 5 is maximum If a ≤ f x ≤ b then a is called the minimum value of f x and b is called the maximum value of f x . f x is the set of all real numbers. The set of output values for a function is called the range of the function. Hence, the range is - 3 ≤ f x ≤ 5 . Enable Ginger Cannot connect to Ginger Check your internet connection or reload the browser Disable in this text field Rephrase Rephrase current sentence Edit in Ginger ×

E M B I B E

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 2 - Functions>EXERCISE 2A

Sue Pemberton Solutions for Chapter: Functions, Exercise 1: EXERCISE 2A

Author:Sue Pemberton

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton Solutions for Chapter: Functions, Exercise 1: EXERCISE 2A

Attempt the practice questions on Chapter 2: Functions, Exercise 1: EXERCISE 2A with hints and solutions to strengthen your understanding. Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book solutions are prepared by Experienced Embibe Experts.