Question

Examine the applicable of MVT for all three functions.

$f\left(x\right)=\left[x\right]forxϵ[5,9]$

$f\left(x\right)=\left[x\right]forxϵ[-2,2]$

$f\left(x\right)=1-x^{2}forxϵ[1,2]$

Answer 1

Here, MVT is not applicable to f(x)=[x] in [5, 9] as f(x) is neither continuous in [5, 9].

($\because$ f is not continuous at 5, 6, 7, 8, 9).

Here, MVT is not applicable to f(x)=[x] in [2, -2] as f(x) is neither continuous in [2, -2].

($\because$ f is not continuous at -2, -1, 0, 1, 2).

Here, $f\left(x\right)=1-x^2$, which is a polynomial function.

Since, a polynomial function is continuous and derivable at all $xϵR$, therefore

(a) f is continuous in [1, 2]

(b) f is derivable in (1, 2) $\Rightarrow$ f satisfies conditions of MVT in [1, 2]

Hence, there exists atleast one real c$ϵ$[1, 2] such that

$f^{\prime }\left(c\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}(\therefore f^{\prime }(c)=-\frac{f\left(b\right)-f\left(a\right)}{b-a})\Rightarrow -2c=\frac{(1-2^2)-(1-1^2)}{2-1}(\because f(x)=1-x^2,\therefore f^{\prime }(x)=-2x)\Rightarrow -2c=-3\Rightarrow c=\frac{3}{2}ϵ(1,2)\therefore MVTisverifiedforthegivenfunctioninthegiveninterval.$