Find the inverse of the function

Mathematics

1 answer:

lozanna [386]2 years ago

4 0

as you already know, to get the inverse of any expression we start off by doing a quick switcheroo on the variables and then solving for "y", let's do so.

$\stackrel{h(x)}{y}~~ = ~~6\sqrt[3]{2x+5}-1\implies \stackrel{\textit{quick switcheroo}}{x~~ = ~~6\sqrt[3]{2y+5}-1} \\\\\\ x+1=6\sqrt[3]{2y+5}\implies \cfrac{x+1}{6}=\sqrt[3]{2y+5}\implies \left( \cfrac{x+1}{6} \right)^3=\left( \sqrt[3]{2y+5} \right)^3$

$\left( \cfrac{x+1}{6} \right)^3=2y+5\implies \left( \cfrac{x+1}{6} \right)^3-5=2y\implies \cfrac{\left( \frac{x+1}{6} \right)^3-5}{2}=y \\\\\\ \cfrac{\left( \frac{x+1}{6} \right)^3}{2}-\cfrac{5}{2}=y\implies \cfrac{~~ \frac{(x+1)^3}{6^3}~~}{2}-\cfrac{5}{2}=y\implies \cfrac{(x+1)^3}{432}-\cfrac{5}{2}=\stackrel{y}{h^{-1}(x)}$

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zheka24 [161]

hmmm first off let's convert the √3 +i to trigonometric form, and then use De Moivre's root theorem, bearing in mind that √3 and i or 1i are both positive, meaning we're on the I Quadrant.

$\bf (\stackrel{a}{\sqrt{3}}~,~\stackrel{b}{1i})\qquad \begin{cases} r=&\sqrt{(\sqrt{3})^2+1^2}\\ &\sqrt{3+1}\\ &2\\ \theta =&tan^{-1}\left( \frac{1}{\sqrt{3}}\right)\\\\ &tan^{-1}\left( \frac{\sqrt{3}}{3} \right)\\ &\frac{\pi }{6} \end{cases}~\hfill \implies ~\hfill 2\left[ cos\left( \frac{\pi }{6}\right) +i~sin\left( \frac{\pi }{6}\right) \right]$

$\bf ~\dotfill\\\\ \qquad \textit{power of two complex numbers} \\\\\ [\quad r[cos(\theta)+isin(\theta)]\quad ]^n\implies r^n[cos(n\cdot \theta)+isin(n\cdot \theta)] \\\\[-0.35em] ~\dotfill$

$\bf \left[ 2\left[ cos\left( \frac{\pi }{6}\right) +i~sin\left( \frac{\pi }{6}\right) \right] \right]^3\implies 2^3\left[ cos\left( 3\cdot \frac{\pi }{6}\right) +i~sin\left( 3\cdot \frac{\pi }{6}\right) \right] \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 8\left[cos\left( \frac{\pi }{2} \right) +i~sin\left( \frac{\pi }{2} \right) \right]~\hfill$