Verify Rolle's theorem for the functionf(x)=x^2-4 x+10 on [0,4] . - Vidyadip

Question

Verify Rolle's theorem for the function$f(x)=x^2-4 x+10 \text { on }[0,4] .$

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Answer

Given that $f(x)=x^2-4 x+10$
$f(x)$ is a polynomial which is continuous on $[0,4]$ and it is differentiable on $(0,4)$.
Let $a=0$, and $b=4$
For $x=a=0$ from (I) we get,
$
f(a)=f(0)=(0)^2-4(0)+10=10
$
For $x=b=4$ from (I) we get,
$
f(b)=f(4)=(4)^2-4(4)+10=10
$
So, here $f(a)=f(b) \quad$ i.e. $f(0)=f(4)=10$
All the conditions of Rolle's theorem are satisfied.
To get the value of $c$, we should have
$f^{\prime}(c)=0$ for some $c \in(0,4)$
Differentiate (I) w.r.t.x.
$f^{\prime}(x)=2 x-4=2(x-4)$
Now, for $x=c$,
$f^{\prime}(c)=0 \Rightarrow 2(c-2)=0 \Rightarrow c=2$
Also $c=2 \in(0,4)$
Thus Rolle's theorem is verified.

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