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) = 2x - x2 on [0, 1]
f(x) = 2x - x2 on [0, 1]
f(x) = 2x - x2
Since a polynomial function is everywhere continuous and differentiable.
Therefore, f(x) is continuous on 0, 1 and differentiable on 0,1
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number $\text{c}\in0,1$ such that
$\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}=\frac{\text{f}(1)-\text{f}(0)}{1}$
Now,
f(x) = 2x - x2
⇒ f'(x) = 2 - 2x,
⇒ f(1) = 2 - 1
⇒ f(1) = 1,
⇒ f(0) = 0
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
$\Rightarrow2-2\text{x}=\frac{1-0}{1}$
$\Rightarrow-2\text{x}=1-2$
$\Rightarrow\text{x}=\frac{1}{2}$
Thus, $\text{c}=\frac{1}{2}\in(1,0)$ such that $\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
Hence, Lagrange's mean value theorem is verified.
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