To find the point on the curve $y = x^{3} - 9 x^{2} + 15 x + 3$ at which the tangents are parallel to the x-axis.

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Solution

$y = x^{3} - q x^{2} + 15 x + 3 \to (i) w h e n tan g e n t a r e p a r a l l e l t o x - a x i s T h e n, d y / d x = 0 d i f f e r e n t i a t e w . r t o x \Rightarrow d y / d x = 3 x^{2} - 18 x + 15 + 0 \Rightarrow d y / d x = 0, 3 x^{2} - 18 x + 15 = 0, x^{2} - 6 x + 5 = 0 \Rightarrow x^{2} - 5 x - x + 5 = 0 x (x - 5) - 1 (x - 5) = 0 a t x = 1 x = 1 o r x = 5 P u t t i n g i n e q u a t i o n (i) y = 1 - 9 \times 1 + 15 \times 1 + 3 = 10 a n d y = 5^{3} - 9 \times 25 + 15 \times 5 + 3 = 22$

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