Question

# If $$y=f(x)$$ is continuous on $$[0,6]$$. differentiable on $$(0,6)$$ $$f(0)=-2$$ and $$f(6)=16$$, then at some point between $$x=0$$ and $$x=6$$, $$f'(x)$$ must be equal to

A
18
B
3
C
3
D
14
E
18

Solution

## The correct option is B $$3$$$$f'(x)=\cfrac { f(b)-f(a) }{ b-a } =\cfrac { 16-(-2) }{ 6-0 } =3$$(Using Lagrange's mean value theorem)Mathematics

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