Question

If `{( x − 5) (x^2 − 2 x + 1)} / {( x − 7) (x^2 + 2 x + 3)}` is positive for all real value of x,show that x has no value between 5 and 7?

Answer 1

`"Please see proof below."`

Explanation:

`"We are given the function:"`

`\qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ { ( x - 5 ) ( x^2 - 2 x + 1 ) } / { ( x - 7 ) ( x^2 + 2 x + 3 ) } \quad.`

`"Before proceeding, let's rewrite the function a little, primarily"`
`"by some factoring:"`

`\qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ { ( x - 5 ) ( x - 1 )^2 } / { ( x - 7 ) ( x^2 + 2 x + 1 + 2 ) }`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ { ( x - 5 ) ( x - 1 )^2 } / { ( x - 7 ) ( ( x + 1 )^2 + 2 ) } \quad.`

`"Thus:"`

`\qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ { ( x - 5 ) ( x - 1 )^2 } / { ( x - 7 ) ( ( x + 1 )^2 + 2 ) } \quad.`

`"So, we may write, in terms that should be clear:"`

`\qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ { ( x - 5 ) cdot "positive" } / { ( x - 7 ) ( "positive" + 2 ) }`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ { ( x - 5 ) cdot "positive" } / { ( x - 7 ) cdot "positive" } \quad.`

`"Thus:"`

`\qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ { ( x - 5 ) cdot "positive" } / { ( x - 7 ) cdot "positive" } \quad.`

`"Now suppose:" \quad 5 < a < 7.`

`"So clearly now:"`

`\qquad \qquad \qquad \qquad \qquad \qquad f(a) \ = \ { ( a - 5 ) cdot "positive" } / { ( a - 7 ) cdot "positive" }`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ { "negative" \ cdot \ "positive" } / { "positive" \ cdot \ "positive" }`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ { "negative" } / { "positive" }`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ "negative" \quad.`

`"So we have shown:"`

`\qquad \qquad \qquad \qquad \qquad \quad 5 < a < 7 \quad => \quad f(a) \ = \ "negative"`

`\qquad "i.e.": \qquad f(a) \ = \ "positive" \quad => \quad 5 < a < 7 \quad \ "is impossible".`

`"And so, finally:"`

`\qquad \qquad \qquad { ( x - 5 ) ( x^2 - 2 x + 1 ) } / { ( x - 7 ) ( x^2 + 2 x + 3 ) } \quad "is positive" \quad =>`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 5 < x < 7 \quad \ "is impossible". \qquad \qquad \ \ square`