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If $f$ is a quintic polynomial which leaves remainder $1$ when divided by $(x-1)^3$, and $-1$ when divided by $(x+1)^3$ , then find the value of first derivative of $f$ at $x=2$.

My approach

Let $$ f = A(x-1)^5 + B(x-1)^4 + C(x-1)^3 +1 $$ Also $$ f = A(x+1)^5 + D(x+1)^4 + E(x+1)^3 -1$$

Is this method correct? Also I want to discuss other methods to solve these kind of problems.

Naweed G. Seldon
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user481779
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  • I would say $(Ax^2 + Bx + C)(x-1)^3 + 1$ and $(Ax^2 + Dx + E)(x+1)^3 - 1$ although there may be a way to go back and forth between the two descritptions – Will Jagy Apr 26 '18 at 17:01
  • It's alright either way, I will leave the comment above. $A(x-1)^2 + B (x-1)+ C = A x^2 + Vx + W$ – Will Jagy Apr 26 '18 at 17:06
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    @WillJagy I have a question. Since $f'$ is going to be a biquadratic polynomial, right? And, $f'$ in the first description of $f$ gives $1$ as a double root, and $f'$ in the second description of $f$ gives $-1$ as a double root. Doesn't that mean, necessarily, $f' = A(x-1)^2(x+1)^2 = A(x^2 - 1)^2$. I don't know how we can calculate $A$, but $f'(2) = 9A$ – Naweed G. Seldon Apr 26 '18 at 17:31
  • Comparing coefficients, we see that $A = 1/5$, therefore $f'(2) = 9/5$ – Naweed G. Seldon Apr 26 '18 at 17:38
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    @junkquill appears you have a good trick I was not looking for. Put an answer... – Will Jagy Apr 26 '18 at 17:42
  • @junkquill which coefficients did you compare? Also I think we can use integration to find the value of A – user481779 Apr 26 '18 at 17:45
  • You're right, I've refined my arguments and have posted them as an answer – Naweed G. Seldon Apr 26 '18 at 18:04
  • If you consider your question to be answered by either Will Jagy or me, please consider accepting one of the answers – Naweed G. Seldon May 14 '18 at 12:58
  • Answer is already accepted – user481779 May 15 '18 at 07:56

2 Answers2

2

Another approach is the extended Euclidean algorithm to find the quintic $f,$ although there is now the possibility of fractions:

$$ \left( x^{3} + 3 x^{2} + 3 x + 1 \right) \left( \frac{ 3 x^{2} - 9 x + 8 }{ 16 } \right) - \left( x^{3} - 3 x^{2} + 3 x - 1 \right) \left( \frac{ 3 x^{2} + 9 x + 8 }{ 16 } \right) = \left( 1 \right) $$

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$$ \left( x^{3} + 3 x^{2} + 3 x + 1 \right) $$

$$ \left( x^{3} - 3 x^{2} + 3 x - 1 \right) $$

$$ \left( x^{3} + 3 x^{2} + 3 x + 1 \right) = \left( x^{3} - 3 x^{2} + 3 x - 1 \right) \cdot \color{magenta}{ \left( 1 \right) } + \left( 6 x^{2} + 2 \right) $$ $$ \left( x^{3} - 3 x^{2} + 3 x - 1 \right) = \left( 6 x^{2} + 2 \right) \cdot \color{magenta}{ \left( \frac{ x - 3 }{ 6 } \right) } + \left( \frac{ 8 x }{ 3 } \right) $$ $$ \left( 6 x^{2} + 2 \right) = \left( \frac{ 8 x }{ 3 } \right) \cdot \color{magenta}{ \left( \frac{ 9 x }{ 4 } \right) } + \left( 2 \right) $$ $$ \left( \frac{ 8 x }{ 3 } \right) = \left( 2 \right) \cdot \color{magenta}{ \left( \frac{ 4 x }{ 3 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( 1 \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ x - 3 }{ 6 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ x + 3 }{ 6 } \right) }{ \left( \frac{ x - 3 }{ 6 } \right) } $$ $$ \color{magenta}{ \left( \frac{ 9 x }{ 4 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 3 x^{2} + 9 x + 8 }{ 8 } \right) }{ \left( \frac{ 3 x^{2} - 9 x + 8 }{ 8 } \right) } $$ $$ \color{magenta}{ \left( \frac{ 4 x }{ 3 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ x^{3} + 3 x^{2} + 3 x + 1 }{ 2 } \right) }{ \left( \frac{ x^{3} - 3 x^{2} + 3 x - 1 }{ 2 } \right) } $$ $$ \left( x^{3} + 3 x^{2} + 3 x + 1 \right) \left( \frac{ 3 x^{2} - 9 x + 8 }{ 16 } \right) - \left( x^{3} - 3 x^{2} + 3 x - 1 \right) \left( \frac{ 3 x^{2} + 9 x + 8 }{ 16 } \right) = \left( 1 \right) $$

Will Jagy
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2

$f$ is our quintic polynomial, and we have the following: $$f=g \cdot (x-1)^3+1 \tag{1}$$ $$f = h \cdot (x+1)^3-1 \tag{2}$$ Where $g$ and $h$ are quadratic polynomials.

Now, since we require $f'(2)$, let's just differentiate these 2 descriptions of $f$, and let's note that since $f$ is quintic, $f'$ is biquadratic. Therefore, we get $$f'=(x-1)^2(g' \cdot (x-1)+3g) \tag{3}$$ $$f'=(x+1)^2(h' \cdot (x+1)+3h) \tag{4}$$ From $(3),(4)$ we see that $1$ and $-1$ are double roots of $f'(x)$. Therefore, necessarily $$f'(x)=K(x-1)^2(x+1)^2=K(x^2-1)^2=K(x^4 - 2x^2 +1) \tag{5}$$ $$\implies f(x)=K\left(\frac{x^5}{5}-\frac{2x^3}{3}+x\right)+C$$ $$(1) \implies 1=f(1)=K\left(\frac{1}{5}-\frac{2}{3}+1\right)+C=\frac{8K}{15}+C \tag{6}$$ $$(2) \implies -1=f(-1)=K\left(-\frac{1}{5}+\frac{2}{3}-1\right)+C=-\frac{8K}{15}+C \tag{7}$$

From $(6),(7)$, we get $$C=0,\ K = \frac{15}{8} \implies f'(x)=\frac{15}{8}(x^2-1)^2, \ \ f(x)=\frac{1}{8}(3x^5 - 10x^3 + 15x)$$

And from $(5)$, we get $$f'(2)=\frac{15\cdot 9}{8}$$

Naweed G. Seldon
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