If {( x − 5) (x^2 − 2 x + 1)} / {( x − 7) (x^2 + 2 x + 3)}(x5)(x22x+1)(x7)(x2+2x+3) is positive for all real value of x,show that x has no value between 5 and 7?

1 Answer
Feb 26, 2018

"Please see proof below." Please see proof below.

Explanation:

"We are given the function:" 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