a) Find the values of x for which the function f(x)=1+4x-x^2 is increasing. b) The graph of y=ax^3 + bx^2 + cx passes through the point (-1, 16) and has a stationary point at (1, -4). Find the values of a, b and c.

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Syakirin: a) Find the values of x for which the function f(x)=1+4x-x^2 is increasing. b) The graph of y=ax^3 + bx^2 + cx passes through the point (-1, 16) and has a stationary point at (1, -4). Find the values of a, b and c.

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August 29, 2014 at 12:56am

ShamSuiisses:b ) 16 = a(-1) ^ 3 + b (-1) ^ 2 + (-1)c 16 = - a + b - c 16 = b - a - c -- (equation1) -4 = a (1) ^ 3 + b (1) ^ 2 + (1)c -4 = a + b + c -- (equation2) (eq1 + eq2) 16 + (-4) = (b - a - c) + (a + b + c) 12 = b + b - a + a - c + c 12 = b + b 2b = 12 b = 6 # (eq1 - eq2) 16 - (-4) = (b - a - c) - (a + b + c) 20 = b - b - a - a - c - c 20 = -2a -2c 20 = -2 (a + c) a + c = -10 -- (equation3)
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- August 29, 2014 at 4:45am
ShamSuiisses:I'm not really sure how to find out 'a' and 'c' though but i know you have to use some information about a stationary point in a graph. If you know what that information is, you can find 'c' then find 'a'.
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- August 29, 2014 at 4:46am
ShamSuiisses:as for part (a).. i dont know. ._.
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- August 29, 2014 at 4:46am
Catherine Ng:[tag]Kho Yen Hong:5208b31e9fc7d[/tag]
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- August 29, 2014 at 11:19am
Little Sky:For part a, dy/dx must be > 0 for a positive gradient. Positive gradient means f(x) is increasing. dy/dx = 4 - 2x dy/dx > 0, 4 - 2x > 0, Therefore, x < 2.
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- August 30, 2014 at 12:22pm
Little Sky:[tag]ShamSuiisses:526cd8712bdbd[/tag]
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- August 30, 2014 at 12:34pm
Kho Yen Hong:a) if the function is increasing, dy/dx >0. dy/dx =4-2x>0, x<2. b) substitute the point to the equation, 16 = a(-1) ^ 3 + b (-1) ^ 2 + (-1)c 16 = - a + b - c --- (1) -4 = a (1) ^ 3 + b (1) ^ 2 + (1)c -4 = a + b + c --- (2) Stationary point implies dy/dx=0 0 = 3a(1)^2 + 2b(1) + c 0 = 3a + 2b + c --- (3) Solve the simultaneous equation gives a=-1, b=6, c=-9
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- October 22, 2014 at 5:15am