Sydney Grammar 2009 help (1 Viewer)

mathpie · Sep 17, 2017

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I've attached the question I don't understand. specifically, I don't understand the 3rd line of the solution, which states there is a double root at the y-coordinates of the stationary points? is that a rule where y-f(x) = 0 has a double root at y=f'(x)??

InteGrand · Sep 17, 2017

mathpie said:
View attachment 34223
View attachment 34222
I've attached the question I don't understand. specifically, I don't understand the 3rd line of the solution, which states there is a double root at the y-coordinates of the stationary points? is that a rule where y-f(x) = 0 has a double root at y=f'(x)??

$\noindent If $f(x)$ is that cubic and $(a,b)$ is the coordinates of a stationary point of it, then $f(a) = b\Rightarrow f(a)-b = 0$ (since the point lies on the graph) and $f'(a) = 0$ (slope $0$ at a stationary point). Hence defining the cubic $C(x):= f(x) -b = x^{3}-3x-b $, we have $C(a) = 0$ and $C'(a) = 0$. Therefore, $a$ is a double root of $C(x)$ from what you know about multiplicity of roots of polynomials. (Of course it can't be a triple root here since $C(x)$ is not like something of the form $(x-a)^{3}$.)$

peter ringout · Sep 18, 2017

Nice question
Just use the formula in a) with p=3 to yield q=plus or minus 2
This means that y=x^3-3x+2 and y=x^3-3x-2 both have double roots....ie max or min situated at the x axis
Sliding y=x^3-3x+2 down 2 units gives a stationary point for y=x^3-3x with a y-value of -2. Similarly for the other

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Sydney Grammar 2009 help (1 Viewer)

mathpie

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InteGrand

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peter ringout

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