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- #1

- Feb 14, 2012

- 3,963

Determine the value of $\dfrac{f(-5)+f(9)}{4}$.

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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,963

Determine the value of $\dfrac{f(-5)+f(9)}{4}$.

- Nov 29, 2013

- 172

Hello.

Determine the value of $\dfrac{f(-5)+f(9)}{4}$.

[tex]f(2)=1654=827*2[/tex]

[tex]f(3)=2481=827*3[/tex]

Therefore:

[tex]\dfrac{f(-5)+f(9)}{4}=\dfrac{827*(-5)+827*(9)}{4}=\dfrac{827(-5+9)}{4}=827[/tex]

Regards

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- #3

- Feb 14, 2012

- 3,963

Good try! But that isn't correct...sorry!Hello.

[tex]f(2)=1654=827*2[/tex]

[tex]f(3)=2481=827*3[/tex]

Therefore:

[tex]\dfrac{f(-5)+f(9)}{4}=\dfrac{827*(-5)+827*(9)}{4}=\dfrac{827(-5+9)}{4}=827[/tex]

Regards

- Mar 31, 2013

- 1,358

Hence f(x) – 827 x = 0 for x = 1, 2, or 3

So f(x) = a(x-b)(x-1)(x-2)(x-3) + 827 x where b is the 4th zero of f(x)- 827 x

Comparing coefficient of $x^4$ we have a = 1

so f(x) = (x-b)(x-1)(x-2)(x-3) + 827 x

rest is as below

f(9) = (9-b) * 8 * 7 * 6 + 827 * 9 = 336(9-b) + 827 * 9

f(-5) = (-5-b) * (-6) *(-7) * (-8) + 827 * (-5) = 336(5+b) + 827 * (-5)

So (f(9) + f(-5))/4 = (336 * 14 + 827 * (4))/4 = 2003

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- #5

- Feb 14, 2012

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Well done,

Hence f(x) – 827 x = 0 for x = 1, 2, or 3

So f(x) = a(x-b)(x-1)(x-2)(x-3) + 827 x where b is the 4th zero of f(x)- 827 x

Comparing coefficient of $x^4$ we have a = 1

so f(x) = (x-b)(x-1)(x-2)(x-3) + 827 x

rest is as below

f(9) = (9-b) * 8 * 7 * 6 + 827 * 9 = 336(9-b) + 827 * 9

f(-5) = (-5-b) * (-6) *(-7) * (-8) + 827 * (-5) = 336(5+b) + 827 * (-5)

So (f(9) + f(-5))/4 = (336 * 14 + 827 * (4))/4 = 2003

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- #6

- Mar 5, 2012

- 9,806

Determine the value of $\dfrac{f(-5)+f(9)}{4}$.

$$g(x)=x^4+Px^3+Qx^2+Rx+S$$

for some real constants $P, Q, R, S$.

It follows that:

$$g(0) = S = 1654$$

$$g(1)+g(-1)=(1 + P + Q + R + S) + (1-P+Q-R+S) = 2+2Q+2S = 2481 +827 = 3308$$

$$Q = \frac{3308 - 2 - 2S}{2} = 1653 - S = 1653 - 1654 = -1$$

$$\dfrac{f(-5)+f(9)}{4} = \dfrac{g(-7)+g(7)}{4}

=\frac 1 4 (2\cdot 7^4 + 2 Q\cdot 7^2 + 2S)

=\frac 1 4 (2\cdot 7^4 - 2 \cdot 7^2 + 2\cdot 1654)

=2003$$

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- #7

- Feb 14, 2012

- 3,963

Bravo,

$$g(x)=x^4+Px^3+Qx^2+Rx+S$$

for some real constants $P, Q, R, S$.

It follows that:

$$g(0) = S = 1654$$

$$g(1)+g(-1)=(1 + P + Q + R + S) + (1-P+Q-R+S) = 2+2Q+2S = 2481 +827 = 3308$$

$$Q = \frac{3308 - 2 - 2S}{2} = 1653 - S = 1653 - 1654 = -1$$

$$\dfrac{f(-5)+f(9)}{4} = \dfrac{g(-7)+g(7)}{4}

=\frac 1 4 (2\cdot 7^4 + 2 Q\cdot 7^2 + 2S)

=\frac 1 4 (2\cdot 7^4 - 2 \cdot 7^2 + 2\cdot 1654)

=2003$$