Find all values of x that satisfies the inequality 2 x - 3 ( x - 2 ) ( x - 4 ) < 0 .

Given that f ( x ) = 2 x - 3 ( x - 2 ) ( x - 4 ) < 0 Equating numerator to zero ⇒ 2 x - 3 = 0 ⇒ x = 3 2 Expression becomes zero at x = 3 2 , but we need to find the value for which expression is less than zero, so we can not take x = 3 2 Equating denominator to zero the function becomes undefined So, x ≠ 2 , x ≠ 4 Now, taking a value in the interval of - ∞ , 3 2 ; let x = 0 f ( 0 ) = 2 ( 0 ) - 3 ( 0 - 2 ) ( 0 - 4 ) = - 3 8 Since, f 0 is negative, therefore, f x < 0 for x ∈ - ∞ , 3 2 Taking a value in the interval of 3 2 , 2 , let x = 7 4 f 7 4 = 2 7 4 - 3 7 4 - 2 7 4 - 4 = 8 9 Since, f 7 4 is positive, therefore f x > 0 for x ∈ 3 2 , 2 Now, taking a value in the interval 2 , 4 , let x = 3 f ( 3 ) = 2 ( 3 ) - 3 ( 3 - 2 ) ( 3 - 4 ) = - 3 Since, f 3 is negative, therefore f x < 0 for x ∈ 2 , 4 Taking a value in the interval 4 , ∞ ;let x = 5 f ( 5 ) = 2 ( 5 ) - 3 ( 5 - 2 ) ( 5 - 4 ) = 7 3 Since, f 5 is positive, therefore f x > 0 for x ∈ 4 , ∞ Thus, f ( x ) < 0 when x ∈ - ∞ , 3 2 ∪ 2 , 4

Solve x 2 - 1 x 2 - 2 x - 15 ≤ 0 .

Given that f ( x ) = x 2 - 4 x 2 - 2 x - 15 ≤ 0 ⇒ x 2 - 4 x 2 - 5 x + 3 x - 15 = ( x - 2 ) ( x + 2 ) x x - 5 + 3 ( x - 5 ) ⇒ ( x - 2 ) ( x + 2 ) ( x - 5 ) ( x + 3 ) ≤ 0 Equating numerator to zero, we get x = 2 , x = - 2 Equating denominator to zero, we get x = 5 , x = - 3 but it makes the f x undefined ∴ x ≠ 5 , x ≠ - 3 So, x = - 3 , x = - 2 , x = 2 , x = 5 are the critical points of f ( x ) Therefore, we have to check the sign of f ( x ) in the intervals - ∞ , - 3 , - 3 , - 2 , - 2 , 2 , 2 , 5 and 5 , ∞ Now, plotting all the critical points on a number line Taking any value of x in the interval - ∞ , - 3 Let, x = - 4 f ( - 4 ) = 16 - 4 ( - 4 - 5 ) ( - 4 + 3 ) = 4 3 Since, f - 4 > 0 , therefore f x > 0 , x ∈ - ∞ , - 3 Taking any value of x in the interval - 3 , - 2 let, x = - 5 2 f - 5 2 = 9 4 - 1 2 15 2 = - 3 5 Since, f - 5 2 < 0 , therefore f x < 0 , x ∈ - 3 , - 2 Taking any value of x in the interval - 2 , 2 At x = 0 , we get f ( 0 ) = 4 15 Since, f 0 > 0 , therefore, f x > 0 , x ∈ - 2 , 2 Taking any value in the interval 2 , 5 At x = 3 , we get f ( 3 ) = 3 2 - 4 ( 3 - 5 ) ( 3 + 3 ) = - 5 18 Since, f 3 < 0 , therefore f x < 0 , x ∈ 2 , 5 Taking any value in the interval 5 , ∞ At x = 6 , we get f ( 6 ) = ( 6 ) 2 - 4 ( 6 - 5 ) ( 6 + 3 ) = 32 9 Since, f 6 > 0 , therefore f x > 0 , x ∈ 5 , ∞ We will also include x = 2 , - 2 , because the given expression becomes zero at these points. Thus, f x ≤ 0 , when x ∈ ( - 3 , - 2 ] ∪ [ 2 , 5 )