y-x-3-3-2x-2-5x-7-find-the-critical-points-of-the-function-

Question Number 12601 by @ANTARES_VY last updated on 26/Apr/17

y = \frac{x^{3}}{3} + 2 x^{2} - 5 x + 7 find the

critical points of the function

Answered by mrW1 last updated on 26/Apr/17

y = \frac{x^{3}}{3} + 2 x^{2} - 5 x + 7 = 0

\Rightarrow x = - 8.16

y^{'} = x^{2} + 4 x - 5 = 0

\Rightarrow (x + 5) (x - 1) = 0

\Rightarrow x = - 5

\Rightarrow x = 1

y^{″} = 2 x + 4 = 0

\Rightarrow x = - 2

Answered by ajfour last updated on 26/Apr/17

\frac{d y}{d x} = x^{2} + 4 x - 5 = {(x + 2)}^{2} - 9

\Rightarrow \frac{d y}{d x} = 0 a t x = - 2 \pm 3 = - 5, 1

l o c a l m a x i m a a t x = - 5

y (- 5) = \frac{121}{3}

l o c a l m i n i m a a t x = 1

y (1) = \frac{13}{3}

\frac{d^{2} y}{{d x}^{2}} = 2 x + 4

\Rightarrow p o i n t o f i n f l e x i o n a t x = - 2

y (- 2) = \frac{67}{3}

y = 0 a t x = α; - 9 < α < - 8

c a n n o t e v a l u a t e .

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