Discussion Forum

Let   $f:[a,b]\rightarrow[1,\infty]$   be a continuous function and g:R → R be defined as $\begin{cases}0 & if & x <a &\\\int_{a}^{x}f(t)dt,  &if& a\leq x\leq b & \\\int_{a}^{b}f(t)dt  &if & x>b\end{cases}$Then,

Answer:

Explanation:

Plan  A function f(x) is continuous at x=a, if 

$\lim_{x \rightarrow a^{-}}$ f(x)= $\lim_{x \rightarrow a^{+}}$ f(x=f(a))

  $\lim_{x \rightarrow a^{-}}\frac{f(x)-f(a)}{x-a}=\lim_{x \rightarrow a^{+}}\frac{f(x)-f(a)}{x-a}$

 i.e, f '(a^-)=f '(a⁺)

 Given that   $f:[a,b]\rightarrow[1,\infty]$

   and     $\begin{cases}0 & if & x <a &\\\int_{a}^{x}f(t)dt,  &if& a\leq x\leq b & \\\int_{a}^{b}f(t)dt  &if & x>b\end{cases}$

 Now, g(a^-)=0=g (a⁺)=g(a)

[as     $g(a^{+})\lim_{x \rightarrow a^{+}}\int_{a}^{x} f(I) dt=0$

  and  $g(a)=\int_{a}^{a} f(t) dt=0$ ]

$g(b^{-})=g(b^{+})=g(b)=\int_{a}^{b}  f(t) dt\Rightarrow$

   is continuous   , $\forall x\epsilon R$

 Now,     $g'(x)=\begin{cases}0 &  & x <a &\\f(x),  && a< x< b & \\0 & & x>b\end{cases}$

 g'(a^-)=0 but g'(a⁺)= $f(a)\geq1$ 

 [  $\because$  Range of f(x)  is   $[1,\infty), \forall x\in[a,b]]$

$\Rightarrow$  g is non-differentiable at x=a

 and g'(b⁺ )=0

 but   $g'(b^{-})=f(b)\geq 1$

$\Rightarrow$  g is not differentiable at x=b