Calculus with Absolute Values

$f(x)$ can be rewritten as follows

$\frac{x}{\left | x^{3} \right |}=\frac{x}{x^{2}\left | x \right |}=\frac{1}{x\left | x \right |}=\frac{1}{x\sqrt{x^{2}}}$

(Using the quotient rule)

$f'(x)=\frac{(x\sqrt{x^{2}}\times 0)-(\frac{\mathrm{d} }{\mathrm{d} x}(x\sqrt{x^{2}}))}{x^{4}}$

$=-\frac{\sqrt{x^{2}}+x(\frac{1}{2}(x^{2})^{-\frac{1}{2}}(2x))}{x^{4}}$

$=-\frac{\sqrt{x^{2}}+\frac{x^{2}}{\sqrt{x^{2}}}}{x^{4}}$

$=-\frac{\frac{2x^{2}}{\sqrt{x^{2}}}}{x^{4}}$

$=-\frac{\frac{2x^{2}}{{\left | x \right |}}}{x^{4}}$

$=-\frac{2}{x^{2}\left | x \right |}$

Finding the negative intersection point of $f(x)$ and $g(x)$

$\frac{1}{x\left | x \right |}=-\frac{2}{x^{2}\left | x \right |}+1$

$\frac{1}{x\left | x \right |}=\frac{x^{2}\left | x \right |-2}{x^{2}\left | x \right |}$

$x^{2}\left | x \right |=x^{5}-2x\left | x \right |$

$x^{4}-x\left | x \right |-2\left | x \right |=0$

$x^{4}-\left | x \right |(x+2)=0$

$x=-1$ (By inspection)

$f(-1)=-1$ so $f(x)$ and $g(x)$ intersect at $(-1,-1)$

Finding the equation of the tangents

$f'(-1)=-2$ (using the previous result for $f'(x)$ )

$y=-2x+c$

using point $(-1,-1)$ , $c=-3 \Rightarrow y=-2x-3$

$g'(-1)=\frac{\mathrm{d} }{\mathrm{d} x}\left [ \frac{-2}{x^{2}\left | x \right |}+1 \right ]_{x=-1}=\frac{\mathrm{d} }{\mathrm{d} x}\left [ \frac{-2}{x^{2}\sqrt{x^{2}}} \right ]_{x=-1}$

(Using the quotient rule)

$\frac{(x^{2}\sqrt{x^{2}}\times 0)-(-2\times \frac{\mathrm{d} }{\mathrm{d} x}(x^{2}\sqrt{x^{2}}))}{x^{6}}$

$=\frac{2(2x\sqrt{x^{2}}+x^{2}(\frac{1}{2}(x^{2})^{-\frac{1}{2}}(2x)))}{x^{6}}$

$=\frac{2(2x\sqrt{x^{2}}+\frac{x^{3}}{\sqrt{x^{2}}})}{x^{6}}$

$=\frac{2(\frac{3x^{3}}{\sqrt{x^{2}}})}{x^{6}}$

$=\frac{\frac{6x^{3}}{\left | x \right |}}{x^{6}}$

$=\frac{6}{x^{3}\left | x \right |}$

$\left [ \frac{6}{x^{3}\left | x \right |} \right ]_{x=-1}=-6$

$y=-6x+c$

using point $(-1,-1)$ , $c=-7 \Rightarrow y=-6x-7$

Finding $A$

$A=\int_{-1}^{0}(-2x-3)-(-6x-7){\mathrm{d} x}$

$A=\int_{-1}^{0}4x+4{\mathrm{d} x}$

$\left [ 2x^{2}+4x \right ]_{-1}^{0}=-(2-4)=\mathbf{2}$