Question

What is the range of the function $f\left( x \right)={{x}^{2}}+2x+2$ ?

Answer 1

To find the range of the function $f\left( x \right)={{x}^{2}}+2x+2$ , we have to rewrite this equation by splitting the constant as combining the terms. We will obtain $\Rightarrow f\left( x \right)=\left( {{x}^{2}}+2x+1 \right)+1$ . Now, we have to use the algebraic identity to get the result $f\left( x \right)={{\left( x+1 \right)}^{2}}+1$ . Now, we have to consider ${{\left( x+1 \right)}^{2}}$ , which will always be positive. Now, we have to make an inequality whose RHS will be the RHS of the function $f\left( x \right)$ . From this inequality, we can find the range of the given function.

Complete step-by-step answer:
We have to find the range of the function $f\left( x \right)={{x}^{2}}+2x+2$ . Let us first split the constant as $1+1$ .
$\Rightarrow f\left( x \right)={{x}^{2}}+2x+1+1$
We can group the terms as follows.
$\Rightarrow f\left( x \right)=\left( {{x}^{2}}+2x+1 \right)+1$
We can see that the terms inside the parenthesis is of the form ${{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}$ , where $a=1$ and $b=1$ . Hence, we can write the above equation as
$\Rightarrow f\left( x \right)={{\left( x+1 \right)}^{2}}+1...\left( i \right)$
We know that ${{\left( x+1 \right)}^{2}}$ will always be positive, that is, greater than or equal to 0, for all $x\in \mathbb{R}$ .
$\Rightarrow {{\left( x+1 \right)}^{2}}\ge 0$
Let us add 1 to both the sides to make the LHS of the above equation similar to equation (i).
$\begin{aligned} & \Rightarrow {{\left( x+1 \right)}^{2}}+1\ge 0+1 \\\ & \Rightarrow {{\left( x+1 \right)}^{2}}+1\ge 1...\left( ii \right) \\\ \end{aligned}$
We know that the range of a function is the spread of possible y-values (minimum y-value to maximum y-value). Hence, we can write the range of the given function from (ii) as
$y\in \left[ 1,\infty \right)$
Therefore, the range of the function $f\left( x \right)={{x}^{2}}+2x+2$ is $\left[ 1,\infty \right)$ .

Note: Students must know algebraic identities to simplify the equations. We have used a closed interval in the range for 1 and open interval for 0 because from the inequality (ii), we obtained a $\ge$ sign. We can also find the range by substituting different real numbers for x and computing the corresponding y values ( f(x) ). We can then interpret the range from the values of y or from the graph. Let us see the graph of the function $f\left( x \right)={{x}^{2}}+2x+2$ , which is a parabola.

We can see that the minimum value of y is -1 and maximum value is infinity.