Question

Verify L M V T for the function f(x)=log x,€ [1,e]​

Answer 1

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᯽ʟᴀɢʀᴀɴɢᴇ ᴍᴇᴀɴ ᴠᴀʟᴜᴇ ᴛʜᴇᴏʀᴇᴍ᯽

If a function $f: [a, b] \to R$ is

• Continuous in [a, b]
• Differentiable in (a, b)

then there exist atleast one $c \in (a, b)$ such that

• $f'(c) = \frac{f(b) - f(a)} {b-a}$

★ᴠᴇʀɪғɪᴄᴀᴛɪᴏɴ★

The given function $f: [1, e] \to R$ is

$\:\:\:\:\:\:\:\:\:\:f(x) = log_{e}(x)$

Continuity in [1, e]

For any $a \in [1, e]$

$\:\:\:\:\:\:\:\:\:\: \lim \limits_{x \to a}f(x) = \lim \limits_{x \to a}log_{e}(x)$

$\implies \lim \limits_{x \to a}f(x)= log_{e}(a)$

$\implies \lim \limits_{x \to a}f(x) = f(a)$

So, f is continuous in $[1, e].$

Differentiability in (1, e)

For any $a \in (1, e)$

$\:\:\:\:\:\:\:\:\:\: f'(a) = \frac{d}{dx}log_{e}(x)|_{x=a}$

$\implies f'(a) = \frac{1 }{x}|_{x=a}$

$\implies f'(a) = \frac{1 }{a}$

So, $f'(a)$ exist at all points in $(1, e).$Hence $f(x)$ is differentiable in $(1, e).$

Both conditions of LMVT are satisfied. Let us check if the results are true.

Now,

$\:\:\:\:\:\:\:\:\:\: f'(c) = \frac{f(b) - f(a)} {b-a}$

$\implies \frac{1 }{c} = \frac{log_{e}(e)- log_{e}(1)} {e-1}$

$\implies \frac{1}{c} = \frac{1- 0} {e-1}$

$\implies c = e-1$

$\implies c ≈ 2.73 - 1$

$\implies c ≈ 1.73$

$\implies c \in (1, e)$

Thus there exist $c \in (1, e)$ such that

$\:\:\:\:\:\:\:\: f'(c) = \frac{f(b) - f(a)} {b-a}$

Hence, LMVT is verified.