f(x) = x^3 + 4x - 8
g(x) = x^7 + 16x^2
f(1) = -3 and f(2) = 8, so f(x) has a zero in 1<x<2.
g(1) = 17 and g(2) = 128+64 = 192 so 17<g(x)<192
given: x^3 = 8-4x = 4(2-x)
substitute for x^3
x^7 = 16x(2-x)^2 = 16x^3 - 64x^2 + 64x
add 16x^2 to both sides
g(x) = x^7 + 16x^2 = 16x^3 - 48x^2 + 64x
substitute for x^3 again
g(x) = 64(2-x) - 48x^2 + 64x
g(x) = 128 - 48x^2
g(x) = 16(8-3x^2)
check:
Using Wolfram Alpha to solve the cubic gives x≈1.3647
So f(1.3647) = 8
Plugging that x value into either the 7th degree polynomial or into the expression 16(8-3x^2) gives the same value: 38.614
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Posted by Larry
on 2024-10-09 09:32:25 |
Solution