If c is a point at which Rolle's theorem holds for the function f(x)=_e(x^2+7x) in the int…

Question

If $c$ is a point at which Rolle's theorem holds for the function $f(x)=\log_e\left(\dfrac{x^2+\alpha}{7x}\right)$ in the interval $[3,4]$, where $\alpha\in\mathbb{R}$, then $f''(c)$ is equal to:

Accepted Answer

For Rolle's theorem to be applicable in [3, 4]

f(3) = f(4)

$$ \Rightarrow $$ $${\log _e}\left( {{{9 + \alpha } \over {21}}} \right) = {\log _e}\left( {{{16 + \alpha } \over {28}}} \right)$$

$$ \Rightarrow $$ $$\left( {{{9 + \alpha } \over {21}}} \right) = \left( {{{16 + \alpha } \over {28}}} \right)$$

$$ \Rightarrow $$ 36 + 4$$\alpha $$ = 48 + 3$$\alpha $$

$$ \Rightarrow $$ $$\alpha $$ = 12

According to Rolle's theorem, f'(c) = 0

where c $$ \in $$ (3, 4)

f'(x) = $${{7x} \over {{x^2} + 12}}$$ $$ \times $$ $$\left( {{{2x\left( {7x} \right) - \left( {{x^2} + 12} \right)7} \over {{7^2}{x^2}}}} \right)$$

$$ \Rightarrow $$ f'(x) = $${{{x^2} - 12} \over {\left( {{x^2} + 12} \right)x}}$$

$$ \therefore $$ f'(c) = $${{{c^2} - 12} \over {\left( {{c^2} + 12} \right)c}}$$

From Rolle's theorem

$${{{c^2} - 12} \over {\left( {{c^2} + 12} \right)c}}$$ = 0

$$ \Rightarrow $$ c2 = 12

Now f''(c) = $${{2c\left( {{c^3} + 12c} \right) - \left( {{c^2} - 12} \right)\left( {3{c^2} + 12} \right)} \over {{{\left( {{c^2} + 12} \right)}^2}{c^2}}}$$

= $${{2\left( {12} \right)\left( {24} \right) - 0} \over {{{\left( {24} \right)}^2} \times 12}}$$

= $${1 \over {12}}$$

Solutions · 1