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

Given:

[latex]f(x) 2x^{2} – 7x + 10 —[1] [/latex]

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

[latex]f(2) – 2(2)^{2} – 7(2) + 10 [/latex]

[latex]\Rightarrow [/latex]2 * 4 – 14 + 10

[latex]\Rightarrow [/latex]8 – 14 + 10

[latex]\Rightarrow [/latex]4

[latex]f(5) – 2(5)^{2} – 7(5) + 10 [/latex]

[latex]\Rightarrow [/latex]2 * 25 – 35 + 10

[latex]\Rightarrow [/latex]50 – 35 + 10

[latex]\Rightarrow [/latex]15 + 10

[latex]\Rightarrow [/latex]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 [latex]\in [/latex](2,5) such that:

[latex]f'(c) = \frac{f(5) – f(2)}{5 – 2} [/latex]

[latex]\Rightarrow 4c – 7 = \frac{25 – 4}{5 – 2} [/latex]

[latex]\Rightarrow 4c = 14 [/latex]

[latex]\Rightarrow c = 3.75 [/latex]

Therefore, c = 3.75 which is [latex] \in [/latex](2,5).

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