`f(x) = x^2, [-2,1]` Determine whether the Mean Value Theorem can be applied to `f` on the closed interval `[a,b]`. If the Mean Value Theorem can be applied, find all values of `c` in the open interval `(a,b)` such that `f'(c) = (f(b) - f(a))/(b - a)`. If the Mean Value Theorem cannot be applied, explain why not.
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The mean value theorem is applicable to the given function, since it is a polynomial function. All polynomial functions are continuous and differentiable on R, hence, the given function is continuous and differentiable on interval.
The mean value theorem states:
`f(b) - f(a) = f'(c)(b-a)`
Replacing 1 for b and -2 for a, yields:
`f(1) - f(-2) = f'(c)(1+ 2)`
Evaluating f(1) and f(-2) yields:
`f(1) = 1^2 => f(1) =1`
`f(-2) = (-2)^2 => f(-2) = 4`
You need to evaluate f'(c):
`f'(c) = (c^2)' => f'(c) = 2c`
Replacing the found values in equation `f(1) - f(-2) = f'(c)(1 + 2):`
`1 - 4 = 2c(1+2) => 6c = -3 => c = -3/6 => c = -1/2 in [-2,1]`
Hence, in this case, the mean value theorem can be applied and the value of c is `c = -1/2` .
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