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3 years ago

Given that g(x)=3x²-5x+7 find the following: g(-x)

1 answer:

5 0

G(x) = 3x² - 5x + 7
g(-x) = -(3x² - 5x + 7)
g(-x) = -3x² + 5x - 7

This equation cannot be solved because of a few reasons,

1. This equation didn't show that it equals to 0.
2. Even if it equals to zero, square root of a negative number cannot be solved.
(I will show you what I mean)

-3x² + 5x - 7
is the same as
3x² - 5x + 7
by shifting the equation,

for example,
1 - 3 = -2
shifting other side
2 = -1 + 3

using 3x² - 5x + 7 to solve,

$Solve\ using\ the\ formular\ \ \boxed{ x= { \frac{-b \pm \sqrt{b^2-4ac} }{2a} } }$
a = 3
b = -5
c = 7
$x= { \frac {-(-5) \pm \sqrt{ (-5)^2-4(3)(7) } }{2(3)} }$
$x= { \frac {5 \pm \sqrt{-59} }{6} }$

∴This equation cannot be solved.

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Substitute these into the given differential equation:

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$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

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$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

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so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

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$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

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