if $\displaystyle f(x)=x+\frac{1}{x}$ find $\displaystyle (f^{-1})'(\frac{17}{4})$
I know that i have to use the formula
$\displaystyle [f^{-1}]'(a)=\frac{1}{f'[f^{-1}(a)]}$
but i am unable to find the inverse.
if $\displaystyle f(x)=x+\frac{1}{x}$ find $\displaystyle (f^{-1})'(\frac{17}{4})$
I know that i have to use the formula
$\displaystyle [f^{-1}]'(a)=\frac{1}{f'[f^{-1}(a)]}$
but i am unable to find the inverse.
$\displaystyle f(x) = x + \frac{1}{x} $
$\displaystyle f'(x) = 1 - \frac{1}{x^{2}} $
so $\displaystyle (f^{-1})'(x+ \frac{1}{x}) = \frac{1}{1 - \frac{1}{x^{2}}} $
notice that $\displaystyle f(4) = 4 + \frac{1}{4} = \frac{17}{4}$
so $\displaystyle (f^{-1})'\Big(\frac{17}{4}\Big) = \frac{1}{1- \frac{1}{16}} = \frac{16}{15}$
Hello, yoman360!
This has nothing to do with derivatives . . .
By inspection, we see that: .$\displaystyle \begin{array}{ccc}f(4) &=& \dfrac{17}{4} \\ \\[-3mm] f\left(\dfrac{1}{4}\right) &=& \dfrac{17}{4} \end{array}$If $\displaystyle f(x)\:=\:x+\frac{1}{x}$. find $\displaystyle f^{-1}\left(\frac{17}{4}\right)$
Therefore: .$\displaystyle f^{-1}\left(\frac{17}{4}\right) \;=\;4,\;\frac{1}{4}$
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If we must find the inverse function first . . .
We have: .$\displaystyle y \;=\;x + \frac{1}{x}$
Switch varoables: .$\displaystyle x \;=\;y + \frac{1}{y}$
Solve for $\displaystyle y\!:\;\;xy \;=\;y^2 + 1 \quad\Rightarrow\quad y^2 - xy + 1 \:=\:0$
Quadratic Formula: .$\displaystyle y \:=\:\frac{x \pm\sqrt{x^2-4}}{2}$
Hence: .$\displaystyle f^{-1}(x) \;=\;\frac{x\pm\sqrt{x^2-4}}{2}$
Therefore: .$\displaystyle f^{-1}\left(\frac{17}{4}\right) \;=\;\frac{\frac{17}{4} \pm\sqrt{\left(\frac{17}{4}\right)^2-4}}{2} \;=\;\frac{\frac{17}{4}\pm\sqrt{\frac{225}{16}}}{2 }$
. . . . . $\displaystyle =\;\frac{\frac{17}{4} \pm\frac{15}{4}}{2} \;=\; \begin{Bmatrix}
\dfrac{\frac{17}{4} + \frac{15}{4}}{2} &=& \dfrac{\frac{32}{4}}{2} &=& \dfrac{8}{2} &=& 4 \\ \\[-3mm]
\dfrac{\frac{17}{4}-\frac{15}{4}}{2} &=& \dfrac{\frac{2}{4}}{2} &=& \dfrac{\frac{1}{2}}{2} &=& \dfrac{1}{4} \end{Bmatrix} $