Answer
A) The derivative of $f\left( x \right)=-3x+7$ at x is $f'\left( x \right)=-3$.
B) The slope of the tangent line to the graph of $f\left( x \right)=-3x+7$ at $x=1$ is $f'\left( 1 \right)=-3$ and at $x=4$ is $f'\left( 4 \right)=-3$.
Work Step by Step
(a)
Consider the function, $f\left( x \right)=-3x+7$.
Now, compute the derivate of $f\left( x \right)=-3x+7$ 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 $f\left( x \right)=-3x+7$.
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ -3\left( x+h \right)+7 \right]-\left( -3x+7 \right)}{h}$
Now, simplify $-3\left( x+h \right)$ by using the distributive property $A\left( B+C \right)=AB+AC$.
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{-3x-3h+7+3x-7}{h}$
Combine the like terms in the numerator.
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{-3h}{h}$
Divide numerator and denominator by h and apply the limits.
$\begin{align}
& f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( -3 \right) \\
& =-3
\end{align}$
Thus, the derivative of $f\left( x \right)=-3x+7$ at x is $f'\left( x \right)=-3$.
(b)
Consider the function, $f\left( x \right)=-3x+7$
From part (a), the derivative of $f\left( x \right)=-3x+7$ at x is $f'\left( x \right)=-3$.
To compute the slope of the tangent line at $x=1$, substitute $x=1$ in $f'\left( x \right)$.
$f'\left( 1 \right)=-3$
To compute the slope of the tangent line at $x=4$, substitute $x=4$ in $f'\left( x \right)$.
$f'\left( 4 \right)=-3$
Thus, the slope of the tangent line to the graph of $f\left( x \right)=-3x+7$ at $x=1$ is $f'\left( 1 \right)=-3$ and at $x=4$ is $f'\left( 4 \right)=-3$
