Verify Lagrange's mean value theorem for the function f(x)=x+4 on the interval [0,5].

Given that f(x)=x+4 The function f(x) is continuous on the closed interval [0,5] and differentiable on the open interval (0,5), so the LMVT is applicable to the function. Differentiate (I) w. r.t.x. f^ (x)=1 2 x+4 Let a=0 and b=5 From (I), & f(a)=f(0)=0+4=2 & f(b)=f(5)=5+4=3 Let c (0,5) such that f^ (c)=f(b)-f(a) b-a 1 2 c+4 =3-2 5-0 =1 5 c+4=5 2 c+4=25 4 c=9 4 (0,5) Thus Lagrange's Mean Value Theorem is verified.

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Question

Verify Lagrange's mean value theorem for the function $f(x)=\sqrt{x+4}$ on the interval $[0,5]$.

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Answer

Given that $f(x)=\sqrt{x+4}$
The function $f(x)$ is continuous on the closed interval $[0,5]$ and differentiable on the open interval $(0,5)$, so the LMVT is applicable to the function.
Differentiate (I) w. r.t.x.
$f^{\prime}(x)=\frac{1}{2 \sqrt{x+4}}$
Let $a=0$ and $b=5$
From (I),
$
\begin{aligned}
& f(a)=f(0)=\sqrt{0+4}=2 \\
& f(b)=f(5)=\sqrt{5+4}=3
\end{aligned}
$
Let $c \in(0,5)$ such that
$
\begin{gathered}
f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \\
\frac{1}{2 \sqrt{c+4}}=\frac{3-2}{5-0}=\frac{1}{5} \\
\therefore \sqrt{c+4}=\frac{5}{2} \Rightarrow c+4=\frac{25}{4} \therefore c=\frac{9}{4} \in(0,5)
\end{gathered}
$
Thus Lagrange's Mean Value Theorem is verified.

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