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