Harder curve sketching question help (1 Viewer)

purplemoon · Dec 4, 2012

hi

Could someone please help me out with this question, i'd appreciate it so much

show that the curve y= (x^2-bc)/(2x-b-c), b>c, has stationary points at x=b and x=c

Carrotsticks · Dec 4, 2012

Trebla · Dec 4, 2012

Carrotsticks said:
$\\ $Using the Quotient Rule, we have $ \frac{dy}{dx} = \frac{2x(2x-b-c) - 2 (x^2-bc)}{(2x-b-c)^2} = 0 $ for stationary point(s).$ \\\\ 2x(2x-b-c) - 2 (x^2-bc) = 0 \\\\ x(2x-b-c) - x^2 + bc = 0 \\\\ 2x^2 - bx - cx - x^2 + bc = 0 \\\\ x^2 - (b+c)x + bc = 0 \\\\ $Clearly, the sum of the roots is $ b+c $ and the product of roots is $ bc $ and thus the roots are $ b $ and $ c. \\\\ $However, if you haven't yet learnt this, you could notice that it factorises to become $ (x-b)(x-c) = 0 $ and therefore the roots of the derivative are $ x=b $ and $ x=c $ and therefore stationary points occur there.$$

I've noticed you're using LaTeX more often in your posts

purplemoon · Dec 5, 2012

thankyou ! i appreciate it

Harder curve sketching question help (1 Viewer)

purplemoon

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Carrotsticks

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Trebla

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