Verify L.M.V. theorem for `f(x)={2+x^3, xlt=1 3x , x >1` on `[-1,2]`

Explanation

Solution:

This theorem states that if `f(x)` be continuous on `[a,b]` and differentiable on `(a,b)` then there exists some c between a and b such that

`f'(c)=[f(b)-f(a)]/[b-a]`

=`3x if x>1 and [a,b]=[−1,2]`

`⟹f ′ (x)=3x^2 if x≤1`

=`3 if x>1`

So, `f'(c)=6-(-2)/2-(-1)`

`⟹3c^ = 5/3`

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