Vieta's Derivatives

Calculus Level 5

f ( x ) = x 3 + x 2 − 3 x + 4 f(x)=x^{3}+x^{2}-3x+4 Let the zeroes of the function above be α , β , γ \alpha,\beta,\gamma . Find f ′ ( α ) f ′ ( β ) f ′ ( γ ) f'(\alpha)f'(\beta)f'(\gamma) .

Inspiration .


The answer is 547.

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Shivam Jadhav by Shivam Jadhav , 22, India

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2 solutions

Chew-Seong Cheong
Jul 22, 2015

Since α \alpha , β \beta and γ \gamma are roots of f ( x ) = x 3 + x 2 − 3 x + 4 f(x) = x^3 + x^2 - 3x + 4 , by Vieta's formulas:

{ α + β + γ = − 1 α β + β γ + γ α = − 3 α β γ = − 4 \begin{cases} \alpha + \beta + \gamma & = -1 \\ \alpha \beta + \beta \gamma + \gamma \alpha & = - 3 \\ \alpha \beta \gamma & = -4 \end{cases}

And since f ′ ( x ) = 3 x 2 + 2 x − 3 f'(x) = 3x^2 + 2x -3 , then we have:

f ′ ( α ) f ′ ( β ) f ′ ( γ ) = ( 3 α 2 + 2 α − 3 ) ( 3 β 2 + 2 β − 3 ) ( 3 γ 2 + 2 γ − 3 ) = 27 α 2 β 2 γ 2 + 18 ( α 2 β 2 γ + α 2 β γ 2 + α β 2 γ 2 ) − 27 ( α 2 β 2 + β 2 γ 2 + γ 2 α 2 ) + 12 ( α 2 β γ + α β 2 γ + α β γ 2 ) − 18 ( α 2 β + β 2 γ + γ 2 α + α β 2 + β γ 2 + γ α 2 ) + 8 α β γ + 27 ( α 2 + β 2 + γ 2 ) − 12 ( α β + β γ + γ α ) + 18 ( α + β + γ ) − 27 = 27 ( α β γ ) 2 + 18 α β γ ( α β + β γ + γ α ) − 27 ( α 2 β 2 + β 2 γ 2 + γ 2 α 2 ) + 12 α β γ ( α + β + γ ) − 18 ( [ α + β + γ ] [ α β + β γ + γ α ] − 3 α β γ ) + 8 α β γ + 27 ( α 2 + β 2 + γ 2 ) − 12 ( α β + β γ + γ α ) + 18 ( α + β + γ ) − 27 [ See Note ] = 27 ( − 4 ) 2 + 18 ( − 4 ) ( − 3 ) − 27 ( 1 ) + 12 ( − 4 ) ( − 1 ) − 18 ( [ − 1 ] [ − 3 ] − 3 [ − 4 ] ) + 8 ( − 4 ) + 27 ( 7 ) − 12 ( − 3 ) + 18 ( − 1 ) − 27 = 547 \begin{aligned} f'(\alpha)f'(\beta)f'(\gamma) & = (3\alpha^2 + 2\alpha -3) (3\beta^2 + 2\beta -3) (3\gamma^2 + 2\gamma -3) \\ & = 27\alpha^2 \beta^2\gamma^2 + 18(\alpha^2 \beta^2\gamma + \alpha^2 \beta\gamma^2 +\alpha\beta^2\gamma^2) \\ & \quad -27(\alpha^2 \beta^2 + \beta^2 \gamma^2 + \gamma^2\alpha^2) + 12(\alpha^2 \beta\gamma+\alpha \beta^2\gamma + \alpha\beta\gamma^2) \\ & \quad -18 (\alpha^2 \beta + \beta^2\gamma + \gamma^2\alpha + \alpha\beta^2 + \beta \gamma^2 + \gamma\alpha^2) \\ & \quad +8\alpha \beta \gamma + 27(\alpha^2 + \beta^2 + \gamma^2) -12(\alpha\beta + \beta\gamma + \gamma\alpha ) \\ & \quad + 18(\alpha + \beta + \gamma) - 27 \\ & = 27(\alpha\beta\gamma)^2 + 18 \alpha \beta \gamma (\alpha\beta + \beta\gamma + \gamma\alpha) \\ & \quad -27(\color{#3D99F6}{\alpha^2 \beta^2 + \beta^2\gamma^2 + \gamma^2 \alpha^2}) + 12\alpha\beta\gamma(\alpha + \beta + \gamma) \\ & \quad -18 ([\alpha + \beta + \gamma][\alpha\beta + \beta \gamma + \gamma\alpha]-3\alpha \beta \gamma) \\ & \quad +8\alpha \beta \gamma + 27(\color{#3D99F6}{\alpha^2 + \beta^2 + \gamma^2}) -12(\alpha\beta + \beta\gamma + \gamma\alpha ) \\ & \quad + 18(\alpha + \beta + \gamma) - 27 \quad \quad \quad \quad \color{#3D99F6} {[\text{See Note}]} \\ & = 27(-4)^2 + 18(-4)(-3) -27(\color{#3D99F6}{1}) + 12(-4)(-1) \\ & \quad -18 ([-1][-3]-3[-4]) +8(-4) + 27(\color{#3D99F6}{7}) -12(-3) \\ & \quad + 18(-1) - 27 \\ & = \boxed{547} \end{aligned}

Note \color{#3D99F6} {\text {Note} }

α 2 + β 2 + γ 2 = ( α + β + γ ) 2 − 2 ( α β + β γ + γ α ) = ( − 1 ) 2 − 2 ( − 3 ) = 7 α 3 + β 3 + γ 3 = ( α + β + γ ) ( α 2 + β 2 + γ 2 ) − ( α β + β γ + γ α ) ( α + β + γ ) + 3 α β γ = ( − 1 ) ( 7 ) − ( − 3 ) ( − 1 ) + 3 ( − 4 ) = − 22 α 4 + β 4 + γ 4 = ( α + β + γ ) ( α 3 + β 3 + γ 3 ) − ( α β + β γ + γ α ) ( α 2 + β 2 + γ 2 ) + α β γ ( α + β + γ ) = ( − 1 ) ( − 22 ) − ( − 3 ) ( 7 ) + ( − 4 ) ( − 1 ) = 47 \begin{aligned} \color{#3D99F6} {\alpha^2 + \beta^2 + \gamma^2} & = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta \gamma + \gamma\alpha) \\ & = (-1)^2 -2(-3) = \color{#3D99F6}{7} \\ \alpha^3 + \beta^3 + \gamma^3 & = (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2) - (\alpha\beta + \beta \gamma + \gamma\alpha)(\alpha + \beta + \gamma) \\ & \quad + 3\alpha \beta \gamma \\ & = (-1)(7)-(-3)(-1)+3(-4) = -22 \\ \color{#D61F06} {\alpha^4 + \beta^4 + \gamma^4} & = (\alpha + \beta + \gamma)(\alpha^3 + \beta^3 + \gamma^3) - (\alpha\beta + \beta \gamma + \gamma\alpha)\\ & \quad (\alpha^2 + \beta^2 + \gamma^2) + \alpha \beta \gamma(\alpha + \beta + \gamma) \\ & = (-1)(-22)-(-3)(7)+(-4)(-1) = \color{#D61F06}{47} \end{aligned}

α 2 β 2 + β 2 γ 2 + γ 2 α 2 = ( α 2 + β 2 + γ 2 ) 2 − ( α 4 + β 4 + γ 4 ) 2 = ( 7 ) 2 − 47 2 = 1 \begin{aligned} \color{#3D99F6}{\alpha^2 \beta^2 + \beta^2 \gamma ^2 + \gamma^2 \alpha^2} & = \frac{(\color{#3D99F6}{\alpha^2 + \beta^2 + \gamma^2})^2 - (\color{#D61F06} {\alpha^4 + \beta^4 + \gamma^4})}{2} \\ & = \frac{ (\color{#3D99F6}{7})^2 - \color{#D61F06}{47}}{2} = \color{#3D99F6}{1} \end{aligned}

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Moderator note:

yes, that's a way of using the Vieta's formula to work through the calculations.

Let us say, x 3 + x 2 − 3 x + 4 = ( x − α ) ( x − β ) ( x − γ ) x^3 + x^2 - 3x + 4 = (x-\alpha)(x-\beta)(x - \gamma)

f ′ ( x ) = ( x − α ) ( x − β ) + ( x − β ) ( x − γ ) + ( x − α ) ( x − γ ) f'(x) = (x-\alpha)(x-\beta) + (x-\beta)(x-\gamma) + (x-\alpha)(x-\gamma)

f ′ ( α ) = ( α − β ) ( α − γ ) f'(\alpha) = (\alpha - \beta)(\alpha - \gamma)

f ′ ( β ) = ( β − α ) ( β − γ ) f'(\beta) = (\beta - \alpha)(\beta - \gamma)

f ′ ( γ ) = ( γ − α ) ( γ − β ) f'(\gamma) = (\gamma - \alpha)(\gamma - \beta)

f ′ ( α ) f ′ ( β ) f ′ ( γ ) = − ( α − β ) 2 ( β − γ ) 2 ( γ − α ) 2 f'(\alpha)f'(\beta)f'(\gamma) = -(\alpha - \beta)^2(\beta - \gamma)^2(\gamma - \alpha)^2

By the definition of discriminant,

Δ = ( α − β ) 2 ( β − γ ) 2 ( γ − α ) 2 \Delta = (\alpha - \beta)^2(\beta - \gamma)^2(\gamma - \alpha)^2

For a cubic polynomial p x 3 + q x 2 + r x + s px^3 + qx^2 + rx + s ,

Δ = q 2 r 2 − 4 p r 3 − 4 q 3 s − 27 p 2 s 2 + 18 p q r s \Delta = q^2r^2 - 4pr^3 - 4q^3s - 27p^2s^2 + 18pqrs

which is equal to − 547 -547 for the given polynomial.

Hence,

f ′ ( α ) f ′ ( β ) f ′ ( γ ) = − ( α − β ) 2 ( β − γ ) 2 ( γ − α ) 2 = − Δ = 547 f'(\alpha)f'(\beta)f'(\gamma) = -(\alpha - \beta)^2(\beta - \gamma)^2(\gamma - \alpha)^2 = -\Delta = \boxed{547}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Since your problem is titled as "vieta", shouldn't you be using the vieta's formula approach instead?

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