Let f : R → R be the functions defined by f(x) = x3 + 5. Then f-1(x) is:
Solution:
we are given that, $\text{f}(\text{x})=\text{x}^3 +5$
Let us suppose, $\text{y}=\text{x}^3+5$
$\Rightarrow\ \text{x}^3=\text{y}-5$
$\Rightarrow\text{x}=(\text{y}-5)^{\frac{1}{3}}$
$\begin{bmatrix}\because\text{f}(\text{x})=\text{y}\\\Rightarrow\text{x}=\text{f}^{-1}(\text{y})\end{bmatrix}$
$\Rightarrow\text{f}^{-3}(\text{y})=(\text{y}-5)^{\frac{1}{3}}$
$\Rightarrow\text{f}^{-1}(\text{x})=(\text{x}-5)^{\frac{1}{3}}$
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where $[x]$ is the greatest integer $\leq x$. Then the value of $\lim _{x \rightarrow 1} g(h(x-1))$ is
$x = 1 + xy\frac{{dy}}{{dx}} + \frac{{{{\left( {xy} \right)}^2}}}{{2!}}{\left( {\frac{{dy}}{{dx}}} \right)^2} + \frac{{{{\left( {xy} \right)}^3}}}{{3!}}{\left( {\frac{{dy}}{{dx}}} \right)^3} + ......$ is