Can the mean value theorum be applied to the function f(x)=1/x^2 on the interval [-2, 1]? explain.

No, it can not be applied.

Step-by-step explanation:

f(x) = 1/x²

f(x) is a polynomial that is not continuous

As,

f(x) = 1/0 is undefines

Secondly, it is not differentiable (i.e. the derivative does not exists on the interval given)

Derivative of this function

f'(x) = (1)x^-2

       = -2x^(-2-1)

       = -2x^(-3)

       = -2/x³

      =  -2/x³

f'(0) = -2/0 is undefined

Thus, mean value theorem can not be applied.

-362

step-by-step explanation:

(12)(4)+2(12)-34(12)+2(12)-3

  \/         \/         \/         \/

  48 +   24 -   408 + 24 - 3

      \   /               \   /

          72     -     432 - 3 = -363

a

step-by-step explanation:

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