If f x = k - 2 x for x ∈ ℝ and g x = 2 x for x ∈ ℝ , x ≠ 0 . Find the value of k for which the equation f g x = x has two equal roots.

Given f x = k - 2 x for x ∈ ℝ and g x = 2 x for x ∈ ℝ , x ≠ 0 We know that f g ( x ) means the function g acts on x first, then f acts on the result. ⇒ f g x = f 2 x ⇒ f g x = k - 2 2 x ⇒ f g x = k - 4 x Consider f g x = x Therefore, k - 4 x = x ⇒ k x - 4 = x 2 ⇒ x 2 - k x + 4 = 0 The quadratic equation a x 2 + b x + c = 0 has two equal roots then b 2 - 4 a c = 0 ⇒ - k 2 - 4 × 1 × 4 = 0 ⇒ k 2 = 16 ⇒ k = ± 4

If f x = k - 2 x for x ∈ ℝ and g x = 2 x for x ∈ ℝ , x ≠ 0 . Find the value of k for which the equation f g x = x has two equal roots.

Given f x = k - 2 x for x ∈ ℝ and g x = 2 x for x ∈ ℝ , x ≠ 0 We know that f g ( x ) means the function g acts on x first, then f acts on the result. ⇒ f g x = f 2 x ⇒ f g x = k - 2 2 x ⇒ f g x = k - 4 x Consider f g x = x Therefore, k - 4 x = x ⇒ k x - 4 = x 2 ⇒ x 2 - k x + 4 = 0 The quadratic equation a x 2 + b x + c = 0 has two equal roots then b 2 - 4 a c = 0 ⇒ - k 2 - 4 × 1 × 4 = 0 ⇒ k 2 = 16 ⇒ k = ± 4

If f x = x 2 - 3 x for x ∈ ℝ and g x = 2 x - 5 for x ∈ ℝ . Find the values of k for which the equation g f x = k has real solutions.

Given f x = x 2 - 3 x for x ∈ ℝ and g x = 2 x - 5 for x ∈ ℝ We know that g f ( x ) means the function f acts on x first, then g acts on the result. ⇒ g f x = g x 2 - 3 x ⇒ g f x = 2 x 2 - 3 x - 5 ⇒ g f x = 2 x 2 - 6 x - 5 Consider g f x = k Therefore, 2 x 2 - 6 x - 5 = k ⇒ 2 x 2 - 6 x - 5 + k = 0 The quadratic equation a x 2 + b x + c = 0 has real roots then b 2 - 4 a c ≥ 0 ⇒ - 6 2 - 4 × 2 × - 5 - k ≥ 0 ⇒ 36 + 40 + 8 k ≥ 0 ⇒ 8 k ≥ 76 ⇒ k ≥ 19 2

If f x = x + 5 2 x - 1 for x ∈ ℝ , x ≠ 1 2 . Show that f f x = x .

If f x = x + 5 2 x - 1 for x ∈ ℝ , x ≠ 1 2 .. We know that g f ( x ) means the function f acts on x first, then g acts on the result. Consider f f x ⇒ f f x = f x + 5 2 x - 1 ⇒ f f x = x + 5 2 x - 1 + 5 2 x + 5 2 x - 1 - 1 ⇒ f f x = x + 5 + 10 x - 5 2 x - 1 2 x + 10 - 2 x + 1 2 x - 1 ⇒ f f x = 11 x 11 ⇒ f f x = x

If f x = 2 x 2 + 4 x - 8 for x ∈ ℝ , x ≥ k . Express 2 x 2 + 4 x - 8 in the form a x + b 2 + c .

Given f ( x ) = 2 x 2 + 4 x - 8 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 + 2 x - 4 Add and subtract the square of half the coefficient of the x -term, ⇒ f x = 2 x 2 + 2 x + 1 2 - 1 2 - 4 Factor the expression as a square of a binomial. ⇒ f x = 2 x + 1 2 - 1 2 - 4 ⇒ f x = 2 x + 1 2 - 5 ⇒ f x = 2 x + 1 2 - 10

If f x = x 2 - 2 x + 4 for x ∈ ℝ . Find the set of value of x for which f x ≥ 7 .

Given f ( x ) = x 2 - 2 x + 4 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 ) = x 2 - 2 x + 4 Add and subtract the square of half the coefficient of the x -term, ⇒ f ( x ) = x 2 - 2 x + - 1 2 - - 1 2 + 4 Factor the expression as a square of a binomial. ⇒ f ( x ) = x - 1 2 - - 1 2 + 4 ⇒ f ( x ) = x - 1 2 - 1 + 4 ⇒ f ( x ) = x - 1 2 + 3 Consider f x ≥ 7 ⇒ x - 1 2 + 3 ≥ 7 ⇒ x - 1 2 - 4 ≥ 0 ⇒ x - 1 2 - 2 2 ≥ 0 ⇒ x - 1 - 2 x - 1 + 2 ≥ 0 ⇒ x - 3 x + 1 ≥ 0 Therefore, x - 3 ≥ 0 or x + 1 ≤ 0 x ≤ - 1 or x ≥ 3

If f x = x 2 - 2 x + 4 for x ∈ ℝ . Express x 2 - 2 x + 4 in the form x - a 2 + b .

Given f ( x ) = x 2 - 2 x + 4 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 ) = x 2 - 2 x + 4 Add and subtract the square of half the coefficient of the x -term, ⇒ f ( x ) = x 2 - 2 x + - 1 2 - - 1 2 + 4 Factor the expression as a square of a binomial. ⇒ f ( x ) = x - 1 2 - - 1 2 + 4 ⇒ f ( x ) = x - 1 2 - 1 + 4 ⇒ f ( x ) = x - 1 2 + 3

If f x = x 2 - 2 x + 4 for x ∈ ℝ . Write down the range of f .

Given f ( x ) = x 2 - 2 x + 4 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 ) = x 2 - 2 x + 4 Add and subtract the square of half the coefficient of the x -term, ⇒ f ( x ) = x 2 - 2 x + - 1 2 - - 1 2 + 4 Factor the expression as a square of a binomial. ⇒ f ( x ) = x - 1 2 - - 1 2 + 4 ⇒ f ( x ) = x - 1 2 - 1 + 4 ⇒ f ( x ) = x - 1 2 + 3 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 = 1 , Therefore, the minimum value of f x occurs when x = 1 When x = 1 then f x = 1 - 1 2 + 3 = 3 Hence, the range is f x ≥ 3

If f x = x 2 - 5 x for x ∈ ℝ and g x = 2 x + 3 for x ∈ ℝ Find f g x .

Given f x = x 2 - 5 x for x ∈ ℝ and g x = 2 x + 3 for x ∈ ℝ We know that f g ( x ) means the function g acts on x first, then f acts on the result. ⇒ f g x = f 2 x + 3 (Since g is a function multiplied by two then added by three) ⇒ f g x = 2 x + 3 2 - 5 2 x + 3