1. Tardigrade
  2. Question
  3. Mathematics
  4. Given: f(x)=4-((1/2)-x)2 / 3 g(x)= begincases( tan [x]/x), x ≠ 0 1 , x=0 endcases h ( x )= x k ( x )=5 log 2( x +3) then in [0,1], Lagranges Mean Value Theorem is NOT applicable to where [ x ] and x denotes the greatest integer and fraction part function.

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Given: $f(x)=4-\left(\frac{1}{2}-x\right)^{2 / 3} g(x)=\begin{cases}\frac{\tan [x]}{x}, & x \neq 0 \\ 1 \quad, & x=0\end{cases}$
$h ( x )=\{ x \} \quad k ( x )=5^{\log _2( x +3)}$ then in $[0,1]$, Lagranges Mean Value Theorem is NOT applicable to
where $[ x ]$ and $\{ x \}$ denotes the greatest integer and fraction part function.

Application of Derivatives

Solution:

f is not differentiable at $x=\frac{1}{2}$
$g$ is not continuous in $[0,1]$ at $x=0$ & 1
$h$ is not continuous in $[0,1]$ at $x=1$
$k ( x )=( x +3)^{\ln _2 5}=( x +3)^{ p }$ where $2< p <3$