Question

How do you find the domain and range of \[f(x) = 8{x^2} - 5x + 2\] ?

Answer 1

Hint: The domain of a function is the complete step of possible values of the independent variable. That is the domain is the set of all possible ‘x’ values which will make the function ‘work’ and will give the output of ‘y’ as a real number. The range of a function is the complete set of all possible resulting values of the dependent variable, after we have substituted the domain.

Complete step by step answer:
Given, \[f(x) = 8{x^2} - 5x + 2\]. Now we can see that for any value of ‘x’ the function \[f(x)\] is well defined. That for \[x \in R\]. That is for \[x = 0\] in \[f(x) = 8{x^2} - 5x + 2\],
\[f(0) = 8{(0)^2} - 5(0) + 2 \\
\Rightarrow f(0)= 2 \\ \]
For \[x = 1\] in \[f(x) = 8{x^2} - 5x + 2\],
\[f(1) = 8{(1)^2} - 5(1) + 2 \\
\Rightarrow f(1) = 8 - 5 + 2 \\
\Rightarrow f(1) = 5 \\ \]
For \[x = - 1\] in \[f(x) = 8{x^2} - 5x + 2\],
\[f( - 1) = 8{( - 1)^2} - 5( - 1) + 2 \\
\Rightarrow f( - 1) = 8 + 5 + 2 \\
\Rightarrow f( - 1) = 15 \\ \]
Similarly for all real numbers for ‘x’ the function \[f(x)\] is well defined.
Hence the domain is \['R'\] . That is all real numbers.

To find the range we need to find the vertex.If we have a quadratic equation \[a{x^2} + bx + c = 0\] then the ‘x’ vertex is given by \[x = - \dfrac{b}{{2a}}\].From the given equation \[8{x^2} - 5x + 2 = 0\] we have \[a = 8\],\[b = - 5\] and \[c = 2\].Then
\[x = - \dfrac{{( - 5)}}{{2(8)}} \\
\Rightarrow x= \dfrac{5}{{16}} \\ \]
To find vertex ‘y’ we substitute ‘x’ value in the above quadratic equation,
\[y = f(x) = 8{x^2} - 5x + 2\]
\[\Rightarrow y = 8{\left( {\dfrac{5}{{16}}} \right)^2} - 5\left( {\dfrac{5}{{16}}} \right) + 2 \\
\Rightarrow y= \left( {8 \times \dfrac{{25}}{{256}}} \right) - \left( {5 \times \dfrac{5}{{16}}} \right) + 2 \\
\Rightarrow y= \dfrac{{25}}{{32}} - \dfrac{{25}}{{16}} + 2 \\ \]
Taking LCM and simplifying we have,
\[ \Rightarrow y= \dfrac{{25 - 50 + 64}}{{32}}\]
\[\Rightarrow y= \dfrac{{89 - 50}}{{32}} \\
\Rightarrow y= \dfrac{{39}}{{32}} \\
\therefore y= 1.219 \\ \]
Thus \[f \geqslant 1.22\].

Hence, the range is \[\{ f \in R:f \geqslant \dfrac{{39}}{{32}}\} \] and the domain is \['R'\].

Note: When finding the domain remember that the denominator of a fraction cannot be zero and the number under a square root sign must be positive in this section. We generally use graphs to find the domain and range. But it is a little bit difficult to draw.