f(x) = x^5−15x^4−10x^2+20 Prove that the equation f(x)=0 has a real solution with x>=1.

Matias Aguirre 2022-07-22 Answered
f ( x) = x 5 15 x 4 10 x 2 + 20
Prove that the equation f ( x ) = 0 has a real solution with x 1.
Idea: Started off by coming up with a boundary [ 1 , x ] for some real number x. f ( 1 ) = 4 so is less than zero, then attempted to find an x that gave me a positive result, very quickly came to the conclusion that any number above 1 when inserted into a function gave me a negative number. Pretty sure I could be missing something really simple. Any help or hints would be appreciated.
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Answers (2)

yermarvg
Answered 2022-07-23 Author has 19 answers
f ( 20 ) is positive. So, check again.
Also, for future purposes, to solve such questions, you can easily check the behaviour of a polynomial at large x by just considering the sign of the highest power variable. If it is positive, then f ( x ) , and if it is negative, then f ( x ) .
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Raynor2i
Answered 2022-07-24 Author has 6 answers

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