## Precalculus (6th Edition) Blitzer

a) The derivative of $f\left( x \right)={{x}^{2}}-8$ at x is $f'\left( x \right)=2x$. b) The slope of the tangent line to the graph of $f\left( x \right)={{x}^{2}}-8$ at $x=-1$ is $f'\left( -1 \right)=-2$ and at $x=3$ is $f'\left( 3 \right)=6$.
(a) Consider the function, $f\left( x \right)={{x}^{2}}-8$ Now, compute the derivate of $f\left( x \right)={{x}^{2}}-8$ 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 $f\left( x \right)={{x}^{2}}-8$. $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ {{\left( x+h \right)}^{2}}-8 \right]-\left( {{x}^{2}}-8 \right)}{h}$ Now, simplify ${{\left( x+h \right)}^{2}}$ by using the property ${{\left( A+B \right)}^{2}}={{A}^{2}}+2AB+{{B}^{2}}$. $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}+2xh+{{h}^{2}}-8-{{x}^{2}}+8}{h}$ Combine the like terms in the numerator; this gives, $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{2xh+{{h}^{2}}}{h}$ Divide numerator and denominator by h. $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 2x+h \right)$ Apply the limits, \begin{align} & f'\left( x \right)=2x+0 \\ & =2x \end{align} Thus, the derivative of $f\left( x \right)={{x}^{2}}-8$ at x is $f'\left( x \right)=2x$. (b) Consider the function, $f\left( x \right)={{x}^{2}}-8$ From part (a), the derivative of $f\left( x \right)={{x}^{2}}-8$ at x is $f'\left( x \right)=2x$. To compute the slope of the tangent line at $x=-1$, substitute $x=-1$ in $f'\left( x \right)$. \begin{align} & f'\left( -1 \right)=2\left( -1 \right) \\ & =-2 \end{align} To compute the slope of the tangent line at $x=3$, substitute $x=3$ in $f'\left( x \right)$. \begin{align} & f'\left( 3 \right)=2\left( 3 \right) \\ & =6 \end{align}