+26381 How to find the inverse of a cubic function?
\(\begin{array}{|lrcll|} \hline 1. & y &=& 2(x-3)^3-16 \\\\ \text{Inverse:} & x &=& 2(y-3)^3-16 \quad & | \quad +16 \\ & x+16 &=& 2(y-3)^3 \quad & | \quad :2 \\ & \frac{x +16}{2}&=& (y-3)^3 \\ & \frac{x}{2} + 8 &=& (y-3)^3 \quad & | \quad \text{cube root both sides} \\ & \sqrt[3]{\frac{x}{2} + 8} &=& y-3 \quad & | \quad +3 \\ & \sqrt[3]{\frac{x}{2} + 8}+3 &=& y \\ & y &=& \sqrt[3]{\frac{x}{2} + 8}+3 \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline 2. & f(x) &=& \sqrt[3]{x + 1}+5 \\\\ \text{Inverse:} & x &=& \sqrt[3]{y + 1}+5 \quad & | \quad -5 \\ & x-5 &=& \sqrt[3]{y + 1} \quad & | \quad \text{cube both sides} \\ & (x-5)^3 &=& y + 1 \quad & | \quad -1 \\ & (x-5)^3-1 &=& y \\ & y &=& (x-5)^3-1 \\ \hline \end{array} \)
\(\begin{array}{|lrcll|} \hline 3. & y &=& \frac12(x-1)^3+3 \\\\ \text{Inverse:} & x &=& \frac12(y-1)^3+3 \quad & | \quad -3 \\ & x-3 &=& \frac12(y-1)^3 \quad & | \quad \cdot 2 \\ & 2(x-3) &=& (y-1)^3 \\ & 2(x-3) &=& (y-1)^3 \quad & | \quad \text{cube root both sides} \\ & \sqrt[3]{2(x-3)} &=& y-1 \quad & | \quad +1 \\ & \sqrt[3]{2(x-3)}+1 &=& y \\ & y &=& \sqrt[3]{2(x-3)}+1 \\ \hline \end{array}\)
+26381 How to find the inverse of a cubic function?
\(\begin{array}{|lrcll|} \hline 1. & y &=& 2(x-3)^3-16 \\\\ \text{Inverse:} & x &=& 2(y-3)^3-16 \quad & | \quad +16 \\ & x+16 &=& 2(y-3)^3 \quad & | \quad :2 \\ & \frac{x +16}{2}&=& (y-3)^3 \\ & \frac{x}{2} + 8 &=& (y-3)^3 \quad & | \quad \text{cube root both sides} \\ & \sqrt[3]{\frac{x}{2} + 8} &=& y-3 \quad & | \quad +3 \\ & \sqrt[3]{\frac{x}{2} + 8}+3 &=& y \\ & y &=& \sqrt[3]{\frac{x}{2} + 8}+3 \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline 2. & f(x) &=& \sqrt[3]{x + 1}+5 \\\\ \text{Inverse:} & x &=& \sqrt[3]{y + 1}+5 \quad & | \quad -5 \\ & x-5 &=& \sqrt[3]{y + 1} \quad & | \quad \text{cube both sides} \\ & (x-5)^3 &=& y + 1 \quad & | \quad -1 \\ & (x-5)^3-1 &=& y \\ & y &=& (x-5)^3-1 \\ \hline \end{array} \)
\(\begin{array}{|lrcll|} \hline 3. & y &=& \frac12(x-1)^3+3 \\\\ \text{Inverse:} & x &=& \frac12(y-1)^3+3 \quad & | \quad -3 \\ & x-3 &=& \frac12(y-1)^3 \quad & | \quad \cdot 2 \\ & 2(x-3) &=& (y-1)^3 \\ & 2(x-3) &=& (y-1)^3 \quad & | \quad \text{cube root both sides} \\ & \sqrt[3]{2(x-3)} &=& y-1 \quad & | \quad +1 \\ & \sqrt[3]{2(x-3)}+1 &=& y \\ & y &=& \sqrt[3]{2(x-3)}+1 \\ \hline \end{array}\)
