f ( x) = x 5 − 15 x 4 − 10 x 2 + 20Prove 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.

Question

f  (  x) =       x    5    −  15      x    4    −  10      x    2    +  20Prove 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.

yermarvg · Accepted Answer

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  )  →  −  ∞.

Raynor2i · Accepted Answer

Hint:   f  (  1  )  &amp;lt;  0 and       lim          x      →      ∞        =  ∞  .

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

Answered question

Answer & Explanation

New Questions in Calculus 2