Verify Lagranges mean value theorem for function `f(x)=x^2-3x+2` on `[-1,\ 2]`

Explanation

Solution:

Given that,

function `f(x) = x2 – 3x + 2, x ∈ [- 2, 3]` Clearly,`f(x)` is continuous in interval `[-2, 3]` and `f'(x)` is finite and exists in `(-2, 3)`. So, f(x) is differentiable in `(- 2, 3)`.


Therefore, f(x) satisfies both conditions of Langrange’s mean value theorem.

So,` f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}`

`\Rightarrow \quad 2 c-3=\frac{f(3)-f(-2)}{3-(-2)}`

`\Rightarrow \quad 2 c-3=\frac{2-12}{5}`

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