Let `f` be a continuous function on `[a ,b]dot` Prove that there exists a number `x in [a , b]` such that `int_a^xf(t)dt=int_x^bf(t)dtdot`

Options

A. `g(x)` is continuous but not differentiable at a
B. `g(x)` is differentiable on `R`
C. `g(x)` is continuous but not differentiable at `b`
D. `g(x)` is continuous and differentiable at either a or b but not both

Answer

A, C

Explanation

For continuity at `x=a`
`lim_(x to a^(+))g(x)=0`
`g(a)=int_(a)^(a)f(t)dt=0`
`lim_(x to a^(+))g(x)=lim_(x to a^(+))int_(a)^(x)f(t)dt=0`
Hence `g(x)` is continuous at `x=a`
For continuity at `x=b`
`lim_(x to b^(-))g(x)=lim_(xto b^(-))int_(a)^(x)f(t)dt=int_(a)^(b)f(t)dt`
`=lim_(x to b^(+))g(x)=g(b)`
Thus, `g(x)` is continuous at `x=b`.
`g'(x)={(0,xlta),(f(x),altxltb),(0,xgtb):}` ltbgt Since `f(x)ge1` for `x epsilon[a,b],g(x)` is non differentiable at `x=a` and `x=b`