## Precalculus (6th Edition) Blitzer

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 $f'\left( 4 \right)=-5$.
(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 $f'\left( 4 \right)=-5$