Question

If \(f\left( x \right) = 3{x^2} - x + 2\), find \(f\left( 2 \right)\), \(f\left( { - 2} \right)\), \(f\left( a \right)\), \(f\left( { - a} \right)\), \(f\left( {a + 1} \right)\), \(2f\left( a \right)\), \(f\left( {2a} \right)\), \(f\left( {{a^2}} \right)\), \({\left( {f\left( a \right)} \right)^2}\), and \(f\left( {a + h} \right)\).

Short answer

The values of \(f\left( 2 \right)\), \(f\left( { - 2} \right)\), \(f\left( a \right)\), \(f\left( { - a} \right)\), \(f\left( {a + 1} \right)\), \(2f\left( a \right)\), \(f\left( {2a} \right)\), \(f\left( {{a^2}} \right)\), \({\left( {f\left( a \right)} \right)^2}\), and \(f\left( {a + h} \right)\) are \(2\), \(16\), \(3{a^2} - a + 2\), \(3{a^2} + a + 2\), \(3{a^2} + 5a + 4\), \(6{a^2} - 2a + 4\), \(12{a^2} - 2a + 2\), \(3{a^4} - {a^2} + 2\), \(9{a^4} - 6{a^3} + 13{a^2} - 4a + 4\), and \(3{a^2} + 6ah + 3{h^2} - a - h + 2\), respectively.

Step-by-step solution

Describe the given information

The given function is,

\(f\left( x \right) = 3{x^2} - x + 2\)

Find \(f\left( 2 \right)\)

Substitute \(x = 2\) in function \(f\left( x \right) = 3{x^2} - x + 2\) to find\(f\left( 2 \right)\).

\(\begin{array}{c}f\left( 2 \right) = 3{\left( 2 \right)^2} - \left( 2 \right) + 2\\ = 12 - 2 + 2\\ = 12\end{array}\)

Therefore, the value of \(f\left( 2 \right)\)is\(2\).

Find \(f\left( { - 2} \right)\)

Substitute \(x = - 2\) in function \(f\left( x \right) = 3{x^2} - x + 2\) to find\(f\left( { - 2} \right)\).

\(\begin{array}{c}f\left( { - 2} \right) = 3{\left( { - 2} \right)^2} - \left( { - 2} \right) + 2\\ = 12 + 2 + 2\\ = 16\end{array}\)

Therefore, the value of \(f\left( { - 2} \right)\)is\(16\).

Find \(f\left( a \right)\)

Substitute \(x = a\) in function \(f\left( x \right) = 3{x^2} - x + 2\) to find\(f\left( a \right)\).

\(\begin{array}{c}f\left( a \right) = 3{\left( a \right)^2} - a + 2\\ = 3{a^2} - a + 2\end{array}\)

Therefore, the value of \(f\left( a \right)\)is\(3{a^2} - a + 2\).

Find \(f\left( { - a} \right)\)

Substitute \(x = - a\) in function \(f\left( x \right) = 3{x^2} - x + 2\) to find\(f\left( { - a} \right)\).

\(\begin{array}{c}f\left( { - a} \right) = 3{\left( { - a} \right)^2} - \left( { - a} \right) + 2\\ = 3{a^2} + a + 2\end{array}\)

Therefore, the value of \(f\left( { - a} \right)\)is\(3{a^2} + a + 2\).

Find \(f\left( {a + 1} \right)\)

Substitute \(x = a + 1\) in function \(f\left( x \right) = 3{x^2} - x + 2\) to find\(f\left( {a + 1} \right)\).

\(\begin{array}{c}f\left( {a + 1} \right) = 3{\left( {a + 1} \right)^2} - \left( {a + 1} \right) + 2\\ = 3\left( {{a^2} + 2a + 1} \right) - \left( {a + 1} \right) + 2\\ = 3{a^2} + 6a + 3 - a - 1 + 2\\ = 3{a^2} + 5a + 4\end{array}\)

Therefore, the value of \(f\left( {a + 1} \right)\)is\(3{a^2} + 5a + 4\).

Find \(2f\left( a \right)\)

Find \(2f\left( a \right)\) as follows.

\(\begin{array}{c}2f\left( a \right) = 2\left( {3{a^2} - a + 2} \right)\\ = 6{a^2} - 2a + 4\end{array}\)

Therefore, the value of \(2f\left( a \right)\)is\(6{a^2} - 2a + 4\).

Find \(f\left( {2a} \right)\)

Substitute \(x = 2a\) in function \(f\left( x \right) = 3{x^2} - x + 2\) to find\(f\left( {2a} \right)\).

\(\begin{array}{c}f\left( {2a} \right) = 3{\left( {2a} \right)^2} - \left( {2a} \right) + 2\\ = 12{a^2} - 2a + 2\end{array}\)

Therefore, the value of \(f\left( {2a} \right)\)is\(12{a^2} - 2a + 2\).

Find \(f\left( {{a^2}} \right)\)

Substitute \(x = {a^2}\) in function \(f\left( x \right) = 3{x^2} - x + 2\) to find\(f\left( {{a^2}} \right)\).

\(\begin{array}{c}f\left( {{a^2}} \right) = 3{\left( {{a^2}} \right)^2} - {a^2} + 2\\ = 3{a^4} - {a^2} + 2\end{array}\)

Therefore, the value of \(f\left( {{a^2}} \right)\)is\(3{a^4} - {a^2} + 2\).

Find \({\left( {f\left( a \right)} \right)^2}\)

Find \({\left( {f\left( a \right)} \right)^2}\) as follows.

\(\begin{array}{c}{\left( {f\left( a \right)} \right)^2} = {\left( {3{a^2} - a + 2} \right)^2}\\ = 9{a^4} - 6{a^3} + 13{a^2} - 4a + 4\end{array}\)

Therefore, the value of \({\left( {f\left( a \right)} \right)^2}\)is\(9{a^4} - 6{a^3} + 13{a^2} - 4a + 4\).

Find \(f\left( {a + h} \right)\)

Substitute \(x = a + h\) in function \(f\left( x \right) = 3{x^2} - x + 2\) to find\(f\left( {a + h} \right)\).

\(\begin{array}{c}f\left( {a + h} \right) = 3{\left( {a + h} \right)^2} - \left( {a + h} \right) + 2\\ = 3\left( {{a^2} + 2ah + {h^2}} \right) - a - h + 2\\ = 3{a^2} + 6ah + 3{h^2} - a - h + 2\\ = 3{a^2} + 6ah + 3{h^2} - a - h + 2\end{array}\)

Therefore, the value of \(f\left( {a + h} \right)\)is\(3{a^2} + 6ah + 3{h^2} - a - h + 2\).