Answer
a) The derivative of $f\left( x \right)={{x}^{2}}-3x+5$ at x is $f'\left( x \right)=2x-3$.
b) The slope of the tangent line to the graph of $f\left( x \right)={{x}^{2}}-3x+5$ at $x=\frac{3}{2}$ is $f'\left( \frac{3}{2} \right)=0$ and at $x=2$ is $f'\left( 2 \right)=1$.
Work Step by Step
(a)
Consider the function, $f\left( x \right)={{x}^{2}}-3x+5$
Now, compute the derivate of $f\left( x \right)={{x}^{2}}-3x+5$ using the formula $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$
To compute $f\left( x+h \right)$, substitute $x=x+h$ in the function $f\left( x \right)={{x}^{2}}-3x+5$.
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ {{\left( x+h \right)}^{2}}-3\left( x+h \right)+5 \right]-\left( {{x}^{2}}-3x+5 \right)}{h}$
Now, simplify ${{\left( x+h \right)}^{2}}$ by using the property ${{\left( A+B \right)}^{2}}={{A}^{2}}+2AB+{{B}^{2}}$ and $-3\left( x+h \right)$ by using the distributive property $A\left( B+C \right)=AB+AC$.
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}+2xh+{{h}^{2}}-3x-3h+5-{{x}^{2}}+3x-5}{h}$
Combine the like terms in the numerator; this gives,
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{2xh+{{h}^{2}}-3h}{h}$
Divide numerator and denominator by h.
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 2x+h-3 \right)$
Apply the limits,
$\begin{align}
& f'\left( x \right)=2x+0-3 \\
& =2x-3
\end{align}$
Thus, the derivative of $f\left( x \right)={{x}^{2}}-3x+5$ at x is $f'\left( x \right)=2x-3$.
(b)
Consider the function, $f\left( x \right)={{x}^{2}}-3x+5$
From part (a), the derivative of $f\left( x \right)={{x}^{2}}-3x+5$ at x is $f'\left( x \right)=2x-3$.
To compute the slope of the tangent line at $x=\frac{3}{2}$, substitute $x=\frac{3}{2}$ in $f'\left( x \right)$.
$\begin{align}
& f'\left( \frac{3}{2} \right)=2\left( \frac{3}{2} \right)-3 \\
& =3-3 \\
& =0
\end{align}$
To compute the slope of the tangent line at $x=2$, substitute $x=2$ in $f'\left( x \right)$.
$\begin{align}
& f'\left( 2 \right)=2\left( 2 \right)-3 \\
& =4-3 \\
& =1
\end{align}$
