If f x = 2 x + 1 for x ∈ ℝ , x ≠ - 1 . Find the values of x for which f x = f f x .

If f x = 2 x + 1 for x ∈ ℝ , x ≠ - 1 .. We know that f f ( x ) means the function f acts on x first, then f acts on the result. Consider f f x ⇒ f f x = f 2 x + 1 ⇒ f f x = 2 2 x + 1 + 1 ⇒ f f x = 2 2 + x + 1 x + 1 ⇒ f f x = 2 x + 1 x + 3 Now, consider f x = f f x ⇒ 2 x + 1 = 2 x + 1 x + 3 ⇒ x + 3 = x + 1 2 ⇒ x + 3 = x 2 + 2 x + 1 ⇒ x 2 + 2 x + 1 - x - 3 = 0 ⇒ x 2 + x - 2 = 0 ⇒ x 2 + 2 x - x - 2 = 0 ⇒ x x + 2 - x + 2 = 0 ⇒ x + 2 x - 1 = 0 ⇒ x = - 2 , 1

Sue Pemberton Solutions for Chapter: Functions, Exercise 2: EXERCISE 2B

Author:Sue Pemberton

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

Attempt the practice questions on Chapter 2: Functions, Exercise 2: EXERCISE 2B 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.

Questions from Sue Pemberton Solutions for Chapter: Functions, Exercise 2: EXERCISE 2B with Hints & Solutions

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 2 - Functions>EXERCISE 2B>Q 18

If $f (x) = \frac{2}{x + 1}$ for $x \in ℝ, x \neq - 1$ .

Find the values of $x$ for which $f (x) = f f (x)$ .

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 2 - Functions>EXERCISE 2B>Q 19

$P (x) = x^{2} - 1$ for $x \in ℝ$

$Q (x) = x + 2$ for $x \in ℝ$

$R (x) = \frac{1}{x}$ for $x \in ℝ, x \neq 0$

$S (x) = \sqrt{x + 1} - 1$ for $x \in ℝ, x \geq - 1$

Function $P, Q, R$ and $S$ are composed in some way to make a new function, $f (x)$ .

For the following, write $f (x)$ in terms of the functions $P, Q, R$ or $S$ ,and state the domain and range for each composite function.

$f (x) = x^{2} + 4 x + 3$

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 2 - Functions>EXERCISE 2B>Q 19

$P (x) = x^{2} - 1$ for $x \in ℝ$

$Q (x) = x + 2$ for $x \in ℝ$

$R (x) = \frac{1}{x}$ for $x \in ℝ, x \neq 0$

$S (x) = \sqrt{x + 1} - 1$ for $x \in ℝ, x \geq - 1$

Function $P, Q, R$ and $S$ are composed in some way to make a new function, $f (x)$ .

For each of the following, write $f (x)$ in terms of the functions $P, Q, R$ or $S$ ,and state the domain and range for each composite function.

$f (x) = x^{2} + 1$

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 2 - Functions>EXERCISE 2B>Q 19

$P (x) = x^{2} - 1$ for $x \in ℝ$

$Q (x) = x + 2$ for $x \in ℝ$

$R (x) = \frac{1}{x}$ for $x \in ℝ, x \neq 0$

$S (x) = \sqrt{x + 1} - 1$ for $x \in ℝ, x \geq - 1$

Function $P, Q, R$ and $S$ are composed in some way to make a new function, $f (x)$ .

For each of the following, write $f (x)$ in terms of the functions $P, Q, R$ or $S$ ,and state the domain and range of composite function.

$f (x) = x$

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 2 - Functions>EXERCISE 2B>Q 19

$P (x) = x^{2} - 1$ for $x \in ℝ$

$Q (x) = x + 2$ for $x \in ℝ$

$R (x) = \frac{1}{x}$ for $x \in ℝ, x \neq 0$

$S (x) = \sqrt{x + 1} - 1$ for $x \in ℝ, x \geq - 1$

Function $P, Q, R$ and $S$ are composed in some way to make a new function, $f (x)$ .

Write $f (x)$ in terms of the functions $P, Q, R$ or $S$ ,and state the domain and range for composite function.

$f (x) = \frac{1}{x^{2}} + 1$

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 2 - Functions>EXERCISE 2B>Q 19

$P (x) = x^{2} - 1$ for $x \in ℝ$

$Q (x) = x + 2$ for $x \in ℝ$

$R (x) = \frac{1}{x}$ for $x \in ℝ, x \neq 0$

$S (x) = \sqrt{x + 1} - 1$ for $x \in ℝ, x \geq - 1$

Function $P, Q, R$ and $S$ are composed in some way to make a new function, $f (x)$ .

Write $f (x)$ in terms of the functions $P, Q, R$ or $S$ ,and state the domain and range for composite function.

$f (x) = \frac{1}{x + 4}$

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 2 - Functions>EXERCISE 2B>Q 19

$P (x) = x^{2} - 1$ for $x \in ℝ$

$Q (x) = x + 2$ for $x \in ℝ$

$R (x) = \frac{1}{x}$ for $x \in ℝ, x \neq 0$

$S (x) = \sqrt{x + 1} - 1$ for $x \in ℝ, x \geq - 1$

Function $P, Q, R$ and $S$ are composed in some way to make a new function, $f (x)$ .

Write $f (x)$ in terms of the functions $P, Q, R$ or $S$ ,and state the domain and range for composite function.

$f (x) = x - 2 \sqrt{x + 1} + 1$

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 2 - Functions>EXERCISE 2B>Q 19

$P (x) = x^{2} - 1$ for $x \in ℝ$

$Q (x) = x + 2$ for $x \in ℝ$

$R (x) = \frac{1}{x}$ for $x \in ℝ, x \neq 0$

$S (x) = \sqrt{x + 1} - 1$ for $x \in ℝ, x \geq - 1$

Function $P, Q, R$ and $S$ are composed in some way to make a new function, $f (x)$ .

Write $f (x)$ in terms of the functions $P, Q, R$ or $S$ ,and state the domain and range for composite function.

$f (x) = x - 1$

Sue Pemberton Solutions for Chapter: Functions, Exercise 2: EXERCISE 2B

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

Questions from Sue Pemberton Solutions for Chapter: Functions, Exercise 2: EXERCISE 2B with Hints & Solutions

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