find-the-local-minimum-and-maximum-value-a-2-y-x-2-a-x-f-x-4-2-x-9-x-3-

Question Number 59788 by aliesam last updated on 14/May/19

f i n d t h e l o c a l m i n i m u m a n d m a x i m u m v a l u e

a^{2} y = x^{2} (a - x)

f (x) = \frac{4}{2 - x} + \frac{9}{x - 3}

Commented by kaivan.ahmadi last updated on 14/May/19

y = \frac{x^{2} (a - x)}{a^{2}} a n d l e t a > 0

y^{'} = \frac{2 a x - 3 x^{2}}{a^{2}} = 0 \Rightarrow x (2 a - 3 x) = 0 \Rightarrow {\begin{cases} x = 0 \\ x = \frac{2 a}{3} \end{cases}

{\begin{cases} x < 0 \Rightarrow f^{'} (x) < 0 \\ 0 < x < \frac{2 a}{3} \Rightarrow f^{'} (x) > 0 \\ x > \frac{2 a}{3} \Rightarrow f^{'} (x) < 0 \end{cases}

\Rightarrow x = 0 i s l o c a l m i n

a n d

x = \frac{2 a}{3} i s l o c a l m a x

i f a < 0 t h e n

x = 0 i s l o c a l m a x a n d x = \frac{2 a}{3} i s l o c a l m i n

Commented by aliesam last updated on 14/May/19

t h a n k y o u s i r i a m w a i t i n g n u m b e r t o s o l u t i o n

Commented by kaivan.ahmadi last updated on 14/May/19

f (x) = \frac{4}{2 - x} + \frac{9}{x - 3} \Rightarrow

D_{f} = R - {2, 3}

f^{'} (x) = \frac{4}{{(2 - x)}^{2}} - \frac{9}{{(x - 3)}^{2}} = 0 \Rightarrow

4 {(x - 3)}^{2} = 9 {(2 - x)}^{2} \Rightarrow

4 x^{2} - 24 x + 36 = 9 x^{2} - 36 x + 36 \Rightarrow

5 x^{2} - 12 x = 0 \Rightarrow x (5 x - 12) = 0 \Rightarrow {\begin{cases} x = 0 \\ x = \frac{12}{5} \end{cases}

x < 0 \Rightarrow f^{'} (x) < 0

0 < x < \frac{12}{5} \Rightarrow f^{'} (x) > 0

x > \frac{12}{5} \Rightarrow f^{'} (x) < 0

\Rightarrow x = 0 m i n a n d x = \frac{12}{5} m a x

Commented by aliesam last updated on 14/May/19

t h a n k y o u s i r

Answered by meme last updated on 14/May/19

y = \frac{x^{2} (a - x)}{a^{2}} . a \neq 0

y^{^{'}} = \frac{2 x (a - x) - x^{2}}{a^{2}} = 0

2 x (a - x) - x^{2} = 0

2 (a - x) = x

2 a = 3 x

x = \frac{2 a}{3}

i f x \in] - \infty; \frac{2 a}{3}] y^{^{'}} ⩽ 0 . y h a s m i n i m a l

i f x \in [\frac{2 a}{3}; + \infty [y^{^{'}} ⩾ 0 . y h a s m a x i m a l

Answered by MJS last updated on 14/May/19

y = - \frac{1}{a^{2}} x^{3} + \frac{1}{a} x^{2}

y^{'} = - \frac{3}{a^{2}} x^{2} + \frac{2}{a} x = 0 \Rightarrow x = 0 \lor x = \frac{2 a}{3}

y^{″} = - \frac{6}{a^{2}} x + \frac{2}{a}

x = 0 \Rightarrow y^{″} = \frac{2}{a} \Rightarrow \max with a < 0; \min with a > 0

x = \frac{2 a}{3} \Rightarrow y^{″} = - \frac{2}{a} \Rightarrow \min with a < 0; \max with a > 0

f (x) = \frac{4}{2 - x} + \frac{9}{x - 3} = \frac{5 x - 6}{(x - 3) (x - 2)}

f^{'} (x) = - \frac{x (5 x - 12)}{{(x - 3)}^{2} {(x - 2)}^{2}} = 0 \Rightarrow x = 0 \lor x = \frac{12}{5}

f^{″} (x) = \frac{2 (5 x^{3} - 18 x^{2} + 36)}{{(x - 3)}^{3} {(x - 2)}^{3}}

f^{″} (0) = \frac{1}{3} > 0 \Rightarrow \min at x = 0

f^{″} (\frac{12}{5}) = - \frac{625}{3} \Rightarrow \max at x = \frac{12}{5}

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