## Range finder

Posted on: October 12th, 2014 by
3

Let $f(x) = \frac{x^2+x+1}{x^2+x+2}$. Show that for real $x$, $3/7\le f(x) \le 1$.

#### 3 Responses to Range finder

x^2+x+1/4 = (x+1/2)^2>=0
x^2+x+2>=1.75 { Adding 1.75 on both sides }
0<=1/(x^2+x+2)= -1/(x^2+x+2)>= -4/7 { Multiplying by -1 }
1>= (x^2+x+1)/(x^2+x+2)>=3/7 { Adding 1 }

Hence proved ðŸ˜›

Some typing error in my last comment

x^2+x+1/4 = (x+1/2)^2>=0
x^2+x+2>=1.75 { Adding 1.75 on both sides }
0<=1/(x^2+x+2)=-1/(x^2+x+2)>= -4/7 { Multiplying by -1 }
1>= (x^2+x+1)/(x^2+x+2)>=3/7 { Adding 1 }

Hence proved ðŸ˜›

Your template is not accepting my answer. gte is greater than or equal to, and lte is less than or equal to, and eq is equals.

x^2+x+1/4 eq (x+1/2)^2 gte 0
x^2+x+2 gte 1.75 { Adding 1.75 on both sides }
0 lte 1/(x^2+x+2) lte 4/7 { Taking reciprocal }
0 gte -1/(x^2+x+2) gte -4/7 { Multiplying by -1 }
1 gte (x^2+x+1)/(x^2+x+2) gte 3/7 { Adding 1 }

Hence proved ðŸ˜›

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