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If $f(x) = \frac{4x+1}{3}$ what is the value of $\left[f^{-1}(1)\right]^{-1}$?

Guest Mar 20, 2018
#1
+20022
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If $f(x) = \frac{4x+1}{3}$ what is the value of $\left[f^{-1}(1)\right]^{-1}$?

$$\begin{array}{|rcll|} \hline f(x) &=& \dfrac{4x+1}{3} \\ \hline y &=& \dfrac{4x+1}{3} \\\\ 3y &=& 4x+1 \\\\ 3y-1 &=& 4x \\\\ x &=& \dfrac{3y-1}{4} \quad & | \quad x \leftrightarrow y \\\\ y &=& \dfrac{3x-1}{4} \\\\ f^{-1}(x) &=& \dfrac{3x-1}{4} \\ \hline f^{-1}(1) &=& \dfrac{3\cdot 1-1}{4} \\\\ &=& \dfrac{2}{4} \\\\ &=& \dfrac{1}{2} \\ \hline \left[f^{-1}(1)\right]^{-1} &=& \left(\dfrac{1}{2}\right)^{-1} \\\\ &=& \dfrac{1}{ \left(\dfrac{1}{2}\right)^{1}} \\\\ &=& \dfrac{1}{ \dfrac{1}{2} } \\\\ &=& 1\cdot \dfrac{2}{1} \\\\ \mathbf{ \left[f^{-1}(1)\right]^{-1}} &\mathbf{=} & \mathbf{2} \\ \hline \end{array}$$

heureka  Mar 20, 2018