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

Step-by-step solution

Obtain the function \(f\left( 2 \right)\)

Substitute 2 for \(x\) in the function \(f\left( x \right)\) to get the function \(f\left( 2 \right)\) as shown below:

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

Thus, \(f\left( 2 \right) = 12\).

Obtain the function \(f\left( { - 2} \right)\)

Substitute \( - 2\) for \(x\) in the function \(f\left( x \right)\) to get the function \(f\left( { - 2} \right)\) as shown below:

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

Thus, \(f\left( { - 2} \right) = 16\).

Obtain the function \(f\left( a \right)\)

Substitute \(a\) for \(x\) in the function \(f\left( x \right)\) to get the function \(f\left( a \right)\) as shown below:

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

Thus, \(f\left( a \right) = 3{a^2} - a + 2\).

Obtain the function \(f\left( { - a} \right)\)

Substitute \( - a\) for \(x\) in the function \(f\left( x \right)\) to get the function \(f\left( { - a} \right)\) as shown below:

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

Thus,\(f\left( { - a} \right) = 3{a^2} + a + 2\).

Obtain the function \(f\left( {a + 1} \right)\)

Substitute \(a + 1\) for \(x\) in the function \(f\left( x \right)\) to get the function \(f\left( {a + 1} \right)\) as shown below:

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

Thus, \(f\left( {a + 1} \right) = 3{a^2} + 5a + 4\).

Obtain the function \(2f\left( a \right)\)

Multiply \(f\left( a \right)\)by 2 and simplify as shown below:

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

Thus, \(2f\left( a \right) = 6{a^2} - 2a + 4\).

Obtain the function \(f\left( {2a} \right)\)

Substitute \(2a\) for \(x\) in the function \(f\left( x \right)\) to get the function \(f\left( {2a} \right)\) as shown below:

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

Thus, \(f\left( {2a} \right) = 12{a^2} - 2a + 2\).

Obtain the function \(f\left( {{a^2}} \right)\) 

Substitute \({a^2}\) for \(x\) in the function \(f\left( x \right)\) to get the function \(f\left( {{a^2}} \right)\) as shown below:

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

Thus, \(f\left( {{a^2}} \right) = 3{a^4} - {a^2} + 2\).

Obtain the function \({\left( {f\left( a \right)} \right)^2}\)

Take the square of \(f\left( a \right)\) and simplify to get the function \({\left( {f\left( a \right)} \right)^2}\) as shown below:

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

Thus, \({\left( {f\left( a \right)} \right)^2} = 9{a^4} - 6{a^3} + 13{a^2} - 4a + 4\).

Obtain the function \(f\left( {a + h} \right)\) 

Substitute \(a + h\) for \(x\) in the function \(f\left( x \right)\) to get the function \(f\left( {a + h} \right)\) as shown below:

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

Thus, \(f\left( {a + h} \right) = 3{a^2} + 3{h^2} + 6ah - a - h + 2\).