# Verify Lagrange's Mean Value Theorem For The Given Function: F(X)= 2x2 - 7x + 10 In [2,5]

Given:

$f(x) 2x^{2} – 7x + 10 —[1]$

Now by substituting the values x = a = 2 and x = b = 5, we get:

$f(2) – 2(2)^{2} – 7(2) + 10$

$\Rightarrow$2 * 4 – 14 + 10

$\Rightarrow$8 – 14 + 10

$\Rightarrow$4

$f(5) – 2(5)^{2} – 7(5) + 10$

$\Rightarrow$2 * 25 – 35 + 10

$\Rightarrow$50 – 35 + 10

$\Rightarrow$15 + 10

$\Rightarrow$25

Now f(2) ≠ f(5)

Since f(x) is a ploynomial function in x, then f(x) is continous in [2,5]

Since f(x) is a polynomial function in x, then it can be differentiated into:

f'(x) = 4 – 7

Then by Lmvt c $\in$(2,5) such that:

$f'(c) = \frac{f(5) – f(2)}{5 – 2}$

$\Rightarrow 4c – 7 = \frac{25 – 4}{5 – 2}$

$\Rightarrow 4c = 14$

$\Rightarrow c = 3.75$

Therefore, c = 3.75 which is $\in$(2,5).

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