## Precalculus (6th Edition) Blitzer

a) The derivative of $f\left( x \right)=3.2{{x}^{2}}+2.1x$ at x is $f'\left( x \right)=6.4x+2.1$. b) The slope of the tangent line to the graph of $f\left( x \right)=3.2{{x}^{2}}+2.1x$ at $x=0$ is $f'\left( 0 \right)=2.1$ and at $x=4$ is $f'\left( 4 \right)=27.7$.
(a) Consider the function, $f\left( x \right)=3.2{{x}^{2}}+2.1x$ Now, compute the derivate of $f\left( x \right)=3.2{{x}^{2}}+2.1x$ 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 $f\left( x \right)=3.2{{x}^{2}}+2.1x$. $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ 3.2{{\left( x+h \right)}^{2}}+2.1\left( x+h \right) \right]-\left( 3.2{{x}^{2}}+2.1x \right)}{h}$ 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{3.2\left( {{x}^{2}}+2xh+{{h}^{2}} \right)+2.1x+2.1h-3.2{{x}^{2}}-2.1x}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{3.2{{x}^{2}}+6.4xh+3.2{{h}^{2}}+2.1x+2.1h-3.2{{x}^{2}}-2.1x}{h} \end{align} Combine the like terms in the numerator; this gives, $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{6.4xh+3.2{{h}^{2}}+2.1h}{h}$ Divide numerator and denominator by h. $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 6.4x+3.2h+2.1 \right)$ Apply the limits, \begin{align} & f'\left( x \right)=6.4x+3.2\left( 0 \right)+2.1 \\ & =6.4x+2.1 \end{align} Thus, the derivative of $f\left( x \right)=3.2{{x}^{2}}+2.1x$ at x is $f'\left( x \right)=6.4x+2.1$. (b) Consider the function, $f\left( x \right)=3.2{{x}^{2}}+2.1x$ From part (a), the derivative of $f\left( x \right)=3.2{{x}^{2}}+2.1x$ at x is $f'\left( x \right)=6.4x+2.1$. 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)=6.4\left( 0 \right)+2.1 \\ & =0+2.1 \\ & =2.1 \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)=6.4\left( 4 \right)+2.1 \\ & =25.6+2.1 \\ & =27.7 \end{align}