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# find the inverse : ((√3x^4-8)/5)+10)9=f(x)

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find the inverse : ((√3x^4-8)/5)+10)9=f(x)

Guest Sep 22, 2014

#1
+10

$$\begin{array}{rll} f(x)&=&9(\frac{\sqrt{3}x^4-8}{5}+10)\\\\ y&=&9(\frac{\sqrt{3}x^4-8}{5}+10)\\\\ \mbox{The inverse function will be}\\\\ x&=&9(\frac{\sqrt{3}y^4-8}{5}+10)\\\\ \frac{x}{9}&=&\frac{\sqrt{3}y^4-8}{5}+10\\\\ \frac{x}{9}-10&=&\frac{\sqrt{3}y^4-8}{5}\\\\ 5\left(\frac{x}{9}-10\right)&=&\sqrt{3}y^4-8\\\\ \frac{5x}{9}-50&=&\sqrt{3}y^4-8\\\\ \frac{5x}{9}-42&=&\sqrt{3}y^4\\\\ \frac{5x-42*9}{9}&=&\sqrt{3}y^4\\\\ \frac{5x-378}{9}&=&\sqrt{3}y^4\\\\ \frac{5x-378}{9\sqrt3}&=&y^4\\\\ \frac{(5x-378)\sqrt3}{27}&=&y^4\\\\ y^4&=&\frac{(5x-378)\sqrt3}{27}\\\\ y&=&\pm\sqrt[4]{\frac{(5x-378)\sqrt3}{27}}\\\\ f^{-1}(x)&=&\pm\sqrt[4]{\frac{(5x-378)\sqrt3}{27}}\\\\ \end{array}$$

That's what I get anyway.

Guest Sep 23, 2014
#1
+10

$$\begin{array}{rll} f(x)&=&9(\frac{\sqrt{3}x^4-8}{5}+10)\\\\ y&=&9(\frac{\sqrt{3}x^4-8}{5}+10)\\\\ \mbox{The inverse function will be}\\\\ x&=&9(\frac{\sqrt{3}y^4-8}{5}+10)\\\\ \frac{x}{9}&=&\frac{\sqrt{3}y^4-8}{5}+10\\\\ \frac{x}{9}-10&=&\frac{\sqrt{3}y^4-8}{5}\\\\ 5\left(\frac{x}{9}-10\right)&=&\sqrt{3}y^4-8\\\\ \frac{5x}{9}-50&=&\sqrt{3}y^4-8\\\\ \frac{5x}{9}-42&=&\sqrt{3}y^4\\\\ \frac{5x-42*9}{9}&=&\sqrt{3}y^4\\\\ \frac{5x-378}{9}&=&\sqrt{3}y^4\\\\ \frac{5x-378}{9\sqrt3}&=&y^4\\\\ \frac{(5x-378)\sqrt3}{27}&=&y^4\\\\ y^4&=&\frac{(5x-378)\sqrt3}{27}\\\\ y&=&\pm\sqrt[4]{\frac{(5x-378)\sqrt3}{27}}\\\\ f^{-1}(x)&=&\pm\sqrt[4]{\frac{(5x-378)\sqrt3}{27}}\\\\ \end{array}$$

That's what I get anyway.

Guest Sep 23, 2014