If the function f(x) = -4e^{left(frac{1-x}{2} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $f(x) = -4e^{\left(\frac{1-x}{2}\right)} + 1 + x + \frac{x^2}{2} + \frac{x^3}{3}$.Since $g(x) = f^{-1}(x)$,we have $g(f(x)) = x$.Differentia

Vedclass · Accepted Answer

$\frac{1}{5}$

Vedclass · Answer

(A) Given $f(x) = -4e^{\left(\frac{1-x}{2}\right)} + 1 + x + \frac{x^2}{2} + \frac{x^3}{3}$.Since $g(x) = f^{-1}(x)$,we have $g(f(x)) = x$.Differentiating both sides with respect to $x$,we get $g'(f(x)) \cdot f'(x) = 1$,which implies $g'(f(x)) = \frac{1}{f'(x)}$.First,we find $x$ such that $f(x) = -\frac{7}{6}$. By inspection,if $x = 1$,then $f(1) = -4e^0 + 1 + 1 + \frac{1}{2} + \frac{1}{3} = -4 + 2 + \frac{3+2}{6} = -2 + \frac{5}{6} = -\frac{7}{6}$.Now,calculate $f'(x) = -4e^{\left(\frac{1-x}{2}\right)} \cdot \left(-\frac{1}{2}\right) + 1 + x + x^2 = 2e^{\left(\frac{1-x}{2}\right)} + 1 + x + x^2$.At $x = 1$,$f'(1) = 2e^0 + 1 + 1 + 1 = 2 + 3 = 5$.Therefore,$g'(-\frac{7}{6}) = g'(f(1)) = \frac{1}{f'(1)} = \frac{1}{5}$.

If the function $f(x) = -4e^{\left(\frac{1-x}{2}\right)} + 1 + x + \frac{x^2}{2} + \frac{x^3}{3}$ and $g(x) = f^{-1}(x)$,then the value of $g'(-\frac{7}{6})$ equals

Similar Questions

Let $S = \{1, 2, 3\}$. Determine whether the function $f: S \rightarrow S$ defined as below has an inverse. Find $f^{-1}$,if it exists: $f = \{(1, 1), (2, 2), (3, 3)\}$.

If $f: R \rightarrow R$ is a mapping defined by $f(x)=x^{3}+5$,then $f^{-1}(x)$ is equal to

If $f:[1, +\infty) \to [2, +\infty)$ is given by $f(x) = x + \frac{1}{x}$,then ${f^{-1}}(x)$ equals:

Let $g$ be the inverse function of $f$ and $f'(x) = \frac{x^{10}}{1 + x^2}$. If $g(2) = a$,then $g'(2)$ is equal to:

Let $f:(2, 3) \to (0, 1)$ be defined by $f(x) = x - [x]$. Then ${f^{ - 1}}(x)$ equals:

Vedclass Test Series

Exam Paper Generator

Online Exam Module