Answer
b)
The slope intercept equation of the tangent line is $y=3x-4$.
Work Step by Step
(b)
Consider the function, $f\left( x \right)={{x}^{3}}-2$
The $x$ coordinate for the tangent is $1$.
Substitute $x=1$ in the function $f\left( x \right)={{x}^{3}}-2$.
$\begin{align}
& f\left( 1 \right)={{\left( 1 \right)}^{3}}-2 \\
& =1-2 \\
& =-1
\end{align}$
Thus, the tangent line passes through $\left( 1,-1 \right)$.
Now, compute the slope of the tangent line for the function $f\left( x \right)={{x}^{3}}-2$ as follows:
The value of $a$ from the point $\left( a,f\left( a \right) \right)$ or $\left( 1,-1 \right)$ is $1$.
Substitute $a=1$ in ${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}$.
${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 1+h \right)-f\left( 1 \right)}{h}$
Substitute $x=1+h$ in the function $f\left( x \right)={{x}^{3}}-2$ and replace $f\left( 1 \right)=-1$.
$\begin{align}
& {{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ {{\left( 1+h \right)}^{3}}-2 \right]-\left[ -1 \right]}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{{{\left( 1+h \right)}^{3}}-2+1}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{{{\left( 1+h \right)}^{3}}-1}{h}
\end{align}$
Use the property
${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}$.
$\begin{align}
& {{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ {{h}^{3}}+1+3{{h}^{2}}+3h \right]-1}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{{{h}^{3}}+3{{h}^{2}}+3h}{h}
\end{align}$
Make the factor of the numerator and simplify.
$\begin{align}
& {{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{h\left( {{h}^{2}}+3h+3 \right)}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\left( {{h}^{2}}+3h+3 \right)
\end{align}$
Apply the limits.
$\begin{align}
& {{m}_{\tan }}={{0}^{2}}+3\left( 0 \right)+3 \\
& =3
\end{align}$
Thus, the slope of the tangent to the graph of $f\left( x \right)={{x}^{3}}-2$ at $\left( 1,-1 \right)$ is $3$.
Substitute ${{x}_{1}}=1$, ${{y}_{1}}=-1$ and $m=3$ in the equation $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$.
$\begin{align}
& y-\left( -1 \right)=3\left( x-1 \right) \\
& y+1=3x-3 \\
& y=3x-3-1 \\
& y=3x-4
\end{align}$
Therefore, the slope intercept equation of the tangent line is $y=3x-4$.
