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I have dervied: $$H_0=3(\frac{\beta-x}{\beta-\alpha})^2-2(\frac{\beta-x}{\beta-\alpha})^3,$$ $$H_1=3(\frac{x-\alpha}{\beta-\alpha})^2-2(\frac{x-\alpha}{\beta-\alpha})^3,$$ $$S_0=(\frac{(\beta-x)^2}{\beta-\alpha})-(\frac{(\beta-x)^3}{(\beta-\alpha)^2}),$$ $$S_1=(-\frac{(x-\alpha)^2}{\beta-\alpha})+(\frac{(x-\alpha)^3}{(\beta-\alpha)^2}),$$ I am supposed to get $$\\ H_0'(\alpha) = 0, \quad H_1'(\alpha) = 0, \quad S_0'(\alpha) = 1,\quad S_1'(\alpha) = 0 \\ H_0'(\beta) = 0, \quad H_1'(\beta) = 0, \quad S_0'(\beta) = 0,\quad S_1'(\beta) = 1 $$ when I take the derivative but when I calculate it for example for H_0 and then plug in [a,b] for x, I get a zero in the denominator both times and that is undefined and incorrect.

Cherry
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  • What do you mean by and then plug in $ [a,b]$ for $x$? – Moo Dec 12 '19 at 18:24
  • I think he means $\alpha$ and $\beta$ – Andrei Dec 12 '19 at 18:24
  • yes sorry in my textbook they are used interchangably, but basically when I plug in first alpha for the derivative of H_0 I get a zero in the denominator which is not giving me the results I am supposed to get. @Andrei – Cherry Dec 12 '19 at 18:26
  • You don't have $x$ in the denominator, so it will always be some power of $(\alpha-\beta)$ – Andrei Dec 12 '19 at 18:27
  • @Cherry: I get $$H_0'(x) = \dfrac{6 (\beta-x)^2}{(\beta-\alpha)^3}-\dfrac{6 (\beta-x)}{(\beta-\alpha)^2}$$ This results in $H_0'(x) = 0$ when substituting $x = \alpha$ or $x = \beta$. Does you derivative work out to that? Fix that and problem will be resolved and you'll have the same issue with all other derivatives. – Moo Dec 12 '19 at 18:28
  • Yes sorry I was plugging in for values of x that were not present. Thank you. – Cherry Dec 12 '19 at 18:34

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$$H'_0(x)=-6\frac{\beta-x}{(\beta-\alpha)^2}+6\frac{(\beta-x)^2}{(\beta-\alpha)^3}$$ If I calculate $H'(\alpha)$ I get: $$H'(\alpha)=-6\frac{\beta-\alpha}{(\beta-\alpha)^2}+6\frac{(\beta-\alpha)^2}{(\beta-\alpha)^3}=0$$ Also, numerators for both terms are zero when $x=\beta$

Andrei
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  • Oh I am so silly, I was plugging in a and b in the denominator despite there not being an x value. Thank you for the help! – Cherry Dec 12 '19 at 18:33