Answer
a)
The derivative of $f\left( x \right)=1.3{{x}^{2}}-1.4x$ at x is $f'\left( x \right)=2.6x-1.4$.
b) The slope of the tangent line to the graph of $f\left( x \right)=1.3{{x}^{2}}-1.4x$ at $x=0$ is $f'\left( 0 \right)=-1.4$ and at $x=4$ is $f'\left( 4 \right)=9$.
Work Step by Step
(a)
Consider the function, $f\left( x \right)=1.3{{x}^{2}}-1.4x$
Now, compute the derivate of $f\left( x \right)=1.3{{x}^{2}}-1.4x$ 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 $f\left( x \right)=1.3{{x}^{2}}-1.4x$.
$f\left( x+h \right)=1.3{{\left( x+h \right)}^{2}}-1.4x$
Now,
$\begin{align}
& f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ 1.3{{\left( x+h \right)}^{2}}-1.4\left( x+h \right) \right]-\left( 1.3{{x}^{2}}-1.4x \right)}{h}
\end{align}$
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 $A\left( B+C \right)=AB+AC$.
$\begin{align}
& f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{1.3\left( {{x}^{2}}+2xh+{{h}^{2}} \right)-1.4x-1.4h-1.3{{x}^{2}}+1.4x}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{1.3{{x}^{2}}+2.6xh+1.3{{h}^{2}}-1.4x-1.4h-1.3{{x}^{2}}+1.4x}{h}
\end{align}$
Combine the like terms in the numerator; this gives,
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{2.6xh+1.3{{h}^{2}}-1.4h}{h}$
Divide numerator and denominator by h.
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 2.6x+1.3h-1.4 \right)$
Apply the limits,
$\begin{align}
& f'\left( x \right)=2.6x+1.3\left( 0 \right)-1.4 \\
& =2.6x+0-1.4 \\
& =2.6x-1.4
\end{align}$
Thus, the derivative of $f\left( x \right)=1.3{{x}^{2}}-1.4x$ at x is $f'\left( x \right)=2.6x-1.4$.
(b)
Consider the function, $f\left( x \right)=1.3{{x}^{2}}-1.4x$
From part (a), the derivative of $f\left( x \right)=1.3{{x}^{2}}-1.4x$ at x is $f'\left( x \right)=2.6x-1.4$.
To compute the slope of the tangent line at $x=0$, substitute $x=0$ in $f'\left( x \right)$.
$\begin{align}
& f'\left( 0 \right)=2.6\left( 0 \right)-1.4 \\
& =0-1.4 \\
& =-1.4
\end{align}$
To compute the slope of the tangent line at $x=4$, substitute $x=4$ in $f'\left( x \right)$.
$\begin{align}
& f'\left( 4 \right)=2.6\left( 4 \right)-1.4 \\
& =10.4-1.4 \\
& =9
\end{align}$
