## Precalculus (6th Edition) Blitzer

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 $f'\left( 4 \right)=-3$.
(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 $f'\left( 4 \right)=-3$