# How so I solve this? 2-(x+1)^(1/2)-x^2+4x-3 , 0<x<1

Jan 14, 2018

$- 2 < 2 - {\left(x + 1\right)}^{\frac{1}{2}} - {x}^{2} + 4 x - 3 < 2 - \sqrt{2}$

#### Explanation:

If you meant what $2 - {\left(x + 1\right)}^{\frac{1}{2}} - {x}^{2} + 4 x - 3$ is between, then this solution is for you.

So, since $x$ is a number greater than 0, we plug in 0 to the expression to get:
$2 - {\left(0 + 1\right)}^{\frac{1}{2}} - {0}^{2} + 4 \left(0\right) - 3$
$2 - {\left(1\right)}^{\frac{1}{2}} - 0 + 4 \left(0\right) - 3$
$2 - 1 - 0 + 0 - 3$
$- 2$
This means that $2 - {\left(x + 1\right)}^{\frac{1}{2}} - {x}^{2} + 4 x - 3$ has to be greater than $- 2$

Similarly, since $x$ is a number smaller than 1, we plug in 1 to the expression to get:
$2 - {\left(1 + 1\right)}^{\frac{1}{2}} - {1}^{2} + 4 \left(1\right) - 3$
$2 - {\left(2\right)}^{\frac{1}{2}} - 1 + 4 \left(1\right) - 3$
$2 - {\left(2\right)}^{\frac{1}{2}} - 1 + 4 - 3$
$2 - \sqrt{2}$
Which is approximately $0.585786437627$. Therefore, $2 - {\left(x + 1\right)}^{\frac{1}{2}} - {x}^{2} + 4 x - 3$ has to be less than $2 - \sqrt{2}$

So we now have :$- 2 < 2 - {\left(x + 1\right)}^{\frac{1}{2}} - {x}^{2} + 4 x - 3 < 2 - \sqrt{2}$