Question
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\text{x}^2-5\text{x}+4\text{ on }[1,4]$

Answer

According to Rolle’s theorem, if f(x) is a real valued function defined on [a, b] such that it is continuous on [a, b], it is differentiable on (a, b) and f(a) = f(b), then there exists a real number $\text{c}\in(\text{a},\text{b})$ such that f(c)= 0.
Now, f(x) is defined for all $\text{x}\in[1,4].$
At each point of [1, 4], the limit of f(x) is equal to the value of the function.
Therefore, f(x) is continuous on [1, 4].
Also, f'(x) = 2x - 5 exists for all $\text{x}\in(1,4).$
So, f(x) is differentiable on (1, 4).
Also,
f(1) = f(4) = 0
Thus, all the three conditions of Rolle’s theorem are satisfied.
Now, we have to show that there exists $\text{c}\in(1,4)$ such that f'(c) = 0.
We have
f'(x) = 2x - 5
$\therefore$ f'(x) = 0
⇒ 2x - 5 = 0
$\Rightarrow\text{x}=\frac{5}{2}$
$\Big[\text{Since }\text{c}=\frac{5}{2}\in(1,4)\text{ such that }\text{f}'(\text{c})=0\Big]$
Hence, Rolle's theorem is verified.

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