Question

according to Rolle's theorem. for a function ( f:[a, b] rightarrow mathbb{R} ) if
(a) ( mathrm{f} ) is continuous in ( [mathrm{a}, mathrm{b}] )
(b) ( f ) is differentiable in ( (a, b) ) ( f(a)=f(b) )
Then there exists some ( c ) in
( (a, b) ) such that ( f^{prime}(c)=0 )
Let's check all conditions:
as ( f(x)=x^{2}+2 x-8 ) is a polynomial function. ( f(x) ) is continuous for all real value of ( x )
hence, ( f(x) ) is continuous in [-4,2]
we know, every polynomial function are differentiable. e.g., ( f(x)= ) ( 2 x+2 )
hence, ( f(x) ) is differentiable in [-4,2]
now, ( f(-4)=(-4)^{2}+2(-4)-8 )
( =16-8-8=16-16=0 )
( f(2)=(2)^{2}+2(2)-8 )
( =4+4-8=8-8=0 )
hence, ( f(-4)=f(2) )
hence, a point c exists in (-4,2) in such that ( f^{prime}(c)=0 ) because ( f^{prime}(x)=2 x+2 )
put ( x=c, f^{prime}(c)=2 c+2=0 )
( c=-1 )
Rolle's theorem is verified

# (b) Verify Rolle's theorem for Cany f(x) = 2x3 + x² - 4x-2

Solution