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anonymous

  • one year ago

A cubic polynomial function f is defined by f(x) = x^3 + ax^2 + b, where a and b are constants. This function has a stationary point at -1. Find a.

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  1. butterflydreamer
    • one year ago
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    Firstly we are given the function: \[f(x) = x^3 + ax^2 + b \] Now what do we know about stationary points? Stationary points occurs when \[f'(x) = 0\] right? So the first step is to differentiate f(x) to find f'(x)

  2. anonymous
    • one year ago
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    f'(x) = 3x^2 + 2x

  3. butterflydreamer
    • one year ago
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    closeee but you forgot "a" .. Since "a" is a constant, when we differentiate ax^2, we get 2ax. \[f(x) = x^3 + ax^2 + b\] \[ f'(x) = 3x^2 + 2ax\] To find stationary points, set f'(x) = 0 So therefore: \[3x^2 + 2ax = 0\] Factorise out the x x(3x + 2a) = 0 Note: x = 0 , -1 (since we're given one of the stationary points is -1) This means that when 3x + 2a = 0, x = -1 Sub it in and solve for a :)!

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