What is the range of the function\[f\left( x \right) = {x^2} + 2x + 2\]?
A.\[\left( {1,\infty } \right)\]
B.\[\left( {2,\infty } \right)\]
C.\[\left( {0,\infty } \right)\]
D.\[\left[ {1,\infty } \right)\]
E.\[\left( { - \infty ,\infty } \right)\]

Hint: In solving the above question, first we will convert the given function by splitting the constant and rewriting the function into a perfect square, then we will find the minimum value of the function and at what value of\[x\] we will get the minimum value, and thus by finding the minimum value and maximum value we will get the desired result.

Formula used: We will use the algebraic identity, i.e., \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\] and the definition of range of function which is given by the all the possible values of output values.

Complete step-by-step solution:
Given \[f\left( x \right) = {x^2} + 2x + 2\],
Now we will convert the quadratic function into vertex form which is given by \[y = a{\left( {x - h} \right)^2} + k\], , then we will get,
\[ \Rightarrow f\left( x \right) = {x^2} + 2x + 1 + 1\]
Now we will write the expression in perfect square, i.e., \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\], by comparing we get, \[a = x\] and \[b = 1\],
Now we will rewrite the equation as,
\[ \Rightarrow f\left( x \right) = {\left( {x + 1} \right)^2} + 1\]
Now from the equation, we can conclude that the value in the square i.e.,\[{\left( {x + 1} \right)^2}\] will always be positive, which means that the value of \[{\left( {x + 1} \right)^2}\] will always be greater than or equal to 0, for all real numbers, i.e., \[x \in R\], which means we can write as,
\[ \Rightarrow {\left( {x + 1} \right)^2} \ge 0\]
Now we will add 1 to both sides we will get,
\[ \Rightarrow {\left( {x + 1} \right)^2} + 1 \ge 0 + 1\]
Now we will simplify we will get,
\[ \Rightarrow {\left( {x + 1} \right)^2} + 1 \ge 1\]
By the definition of the range of the function i.e., the range of the function is given by the all the possible values of output values,i.e., minimum value to the maximum value,
So, now from the above equation, we will get that the range of the function is \[\left[ {1,\infty } \right)\].
Hence the range of the given function i.e.,\[f\left( x \right) = {x^2} + 2x + 2\] will be equal to \[\left[ {1,\infty } \right)\]

The correct option is D.

Note: In this question we have to find the range of the function. Here we will focus on the minimum value of \[f(x)\]. So rewrite the function as a sum of a square of an expression and a constant. As we know the minimum value of a square must be greater than or equal to zero. Using this concept we will make inequality and solve it to find the range.