## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 11 - Section 11.4 - Introduction to Derrivatives - Exercise Set - Page 1174: 21

#### Answer

a) The derivative of $f\left( x \right)={{x}^{3}}+2$ at x is $f'\left( x \right)=3{{x}^{2}}$. b) The slope of the tangent line to the graph of $f\left( x \right)={{x}^{3}}+2$ at $x=-1$ is $f'\left( -1 \right)=3$ and at $x=1$ is $f'\left( 1 \right)=3$.

#### Work Step by Step

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

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