Show-that-the-curve-y-ln-5-7x-8-x-has-no-stationary-point-for-all-real-values-of-x-

Question Number 8244 by lepan last updated on 04/Oct/16

S h o w t h a t t h e c u r v e y = l n (\frac{5 - 7 x}{8 + x}) h a s

n o s t a t i o n a r y p o i n t f o r a l l r e a l v a l u e s

o f x .

Answered by 123456 last updated on 06/Oct/16

y = \ln \frac{5 - 7 x}{8 + x}

\frac{d y}{d x} = \frac{d}{d x} \ln \frac{5 - 7 x}{8 + x} u = \frac{5 - 7 x}{8 + x}

= \frac{d \ln u}{d x}

= \frac{d \ln u}{d u} \cdot \frac{d u}{d x}

= \frac{1}{u} \cdot \frac{d u}{d x}

= \frac{8 + x}{5 - 7 x} \cdot \frac{d}{d x} \frac{5 - 7 x}{8 + x} {\begin{cases} v = 5 - 7 x \\ w = 8 + x \end{cases}

= \frac{8 + x}{5 - 7 x} \cdot \frac{d}{d x} \frac{v}{w} \frac{d}{d x} \frac{u}{v} = \frac{d}{d x} (u \cdot \frac{1}{v}) = \frac{d u}{d x} \cdot \frac{1}{v} + u \cdot \frac{d}{d x} \frac{1}{v} = \frac{d u}{d x} \cdot \frac{1}{v} - u \cdot \frac{d v / x}{v^{2}} = \frac{u^{'} v - {u v}^{'}}{v^{2}}

= \frac{8 + x}{5 - 7 x} \cdot \frac{\frac{d v}{d x} \cdot w - v \cdot \frac{d w}{d x}}{w^{2}}

= \frac{8 + x}{5 - 7 x} \cdot \frac{\frac{d (5 - 7 x)}{d x} \cdot (8 + x) - (5 - 7 x) \cdot \frac{d (8 + x)}{d x}}{{(8 + x)}^{2}}

= \frac{1}{5 - 7 x} \cdot \frac{- 7 \cdot (8 + x) - (5 - 7 x) \cdot 1}{8 + x}

= \frac{- 7 \cdot 8 - 7 x - 5 + 7 x}{(5 - 7 x) (8 + x)}

= - \frac{61}{(5 - 7 x) (8 + x)}

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