Question
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
f(x) = x2 - 2x + 4 on [1, 5]
f(x) = x2 - 2x + 4 on [1, 5]
f(x) = x2 - 2x + 4
Since a polynomial function is everywhere continuous and differentiable.
Therefore, f(x) is continuous on [1, 5] and differentiable on (1, 5).
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number $\text{c}\in(1,5)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(5)-\text{f}(-1)}{5-1}=\frac{\text{f}(5)-\text{f}(-1)}{4}$
Now, f(x) = x2 - 2x + 4
⇒ f'(x) = 2x - 2
⇒ f(5) = 25 - 10 + 4 = 19
⇒ f(1) = 1 - 2 + 4 = 3
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(5)-\text{f}(-1)}{4}$
$\Rightarrow2\text{x}-2=\frac{19-3}{4}$
$\Rightarrow2\text{x}-2-4=0$
$\Rightarrow\text{x}=\frac{6}{2}=3$
Thus, $\text{c}=3\in(1,5)$ such that $\text{f}'(\text{c})=\frac{\text{f}(5)-\text{f}(-1)}{5-1}$
Hence, Lagrange's mean value theorem is verified.
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Find the maximum and minimum value of 2x + y subject to the constraints:
$\text{x}+3\text{y}\geq6,\text{x}-3\text{y}\leq3,3\text{x}+4\text{y}\leq24,$ $-3\text{x}+2\text{y}\leq6,5\text{x}+\text{y}\geq5,\text{x},\text{y}\geq0$