Single Variable Calculus, Chapter 1, 1.1, Section 1.1, Problem 21

The function $f(x) = 3x^2 - x +2$ is given. Evaluate the following:



$f(2),f(-2),f(a),f(-a),f(a+1),2f(a),f(2a),f(a^3),[f(a)]^2$, and $f(a+h)$




$
\begin{equation}
\begin{aligned}
@ f(2) &\\
f(2) =& 3(2)^2 - (2) + 2 &&(\text{Substitute 2 to the function f})
\end{aligned}
\end{equation}
$




$\fbox{$f(2) = 12$}$



$
\begin{equation}
\begin{aligned}

@ f(-2) &\\
f(-2) =& 3(-2)^2 - (-2) + 2 &&(\text{ Substitute -2 to the function f})
\end{aligned}
\end{equation}
$


$\fbox{$f(-2) = 16$}$



$
\begin{equation}
\begin{aligned}
@f(a)&\\
f(a) =& 3(a)^2 - (a)+2 &&(\text{ Substitute a to the function f})
\end{aligned}
\end{equation}
$



$\fbox{$f(a) = 3a^2 - a+2$}$


$
\begin{equation}
\begin{aligned}
@f(-a)&\\

f(-a) =& 3(-a)^2 - (-a) +2 &&(\text{ Substitute -a to the function f})
\end{aligned}
\end{equation}
$



$\fbox{$f(-a) = 3a^2 + a + 2$}$

$
\begin{equation}
\begin{aligned}


@f(a +1)&\\

f(a+1) = &3(a+1)^2 - (a+1)+2 && (\text{ Substitute (a + 1) to the function f})\\


f(a +1) =& 3(a^2+2a+1) - (a+1)+2 && (\text{ Simplify the equation})\\

f(a + 1) =& 3a^2 + 6a + 3- a - 1 + 2 && (\text{Combine like terms})

\end{aligned}
\end{equation}
$



$\fbox{$f(a + 1)= 3a^2 + 5a + 4$}$

$
\begin{equation}
\begin{aligned}


@2f(a)&\\

2f(a) =& 2[3(a)^2 - (a) + 2] &&(\text{ Substitute a to the function } f, \text{ then multiply the answer by 2})

\end{aligned}
\end{equation}
$



$\fbox{$2f(a) = 6a^2 - 2a+4$}$


$
\begin{equation}
\begin{aligned}

@f(2a)&\\

f(2a) =& 3(2a)^2 - (2a) + 2 &&(\text{ Substitute } 2a \text{ to the function} f)
\end{aligned}
\end{equation}
$



$\fbox{$f(2a) = 12a^2 - 2a +2$}$



$
\begin{equation}
\begin{aligned}

@f(a^2)&\\

f(a^2) =& 3(a^2)^2 - (a^2) + 2 && (\text{ Substitute } a^2 \text{ to the function f, then simplify})
\end{aligned}
\end{equation}
$



$\fbox{$f(a^2)= 3a^4 - a^2 + 2$}$





$
\begin{equation}
\begin{aligned}

@[f(a)]^2 &\\

[f(a)]^2 =& [3(a)^2 - (a) + 2]^2 &&(\text{ Substitute a to the function f, then square the answer})\\



[f(a)]^2 =& (3a^2 - a + 2)^2 &&(\text{ Expand the equation})\\



[f(a)]^2 =& 9a^4 - 3a^3 + 6a^2 - 3a^3 + a^2 - 2a + 6a^2 - 2a +4 &&(\text{ Combine like terms})
\end{aligned}
\end{equation}
$



$\fbox{$[f(a)]^2 = 9a^4 - 6a^3 +13a^2 -4a+4$}$



$
\begin{equation}
\begin{aligned}

@f(a + h) &\\

f(a+h) =& 3(a+h)^2 - (a+h)+2 && (\text{ Substitute } f( a + h) \text{ to the function} f)\\



f(a + h)=& 3(a^2 + 2ah +h^2) - a - h +2
\end{aligned}
\end{equation}
$



$\fbox{$f(a + h) = 3a^2 + 6ah +3h^2-a-h+2$}$