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f(x) = 11x^4 - 3x^3 - 10x^2 + 9x + 18
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You need to test if the given polynomial function has any zeroes, hence, you may perform the derivative test, such that:
`f'(x) = 44x^3 - 9x^2 - 20x + 9`
Evaluating the zeroes of the function ` f'(x)` yields:
`44x^3 - 9x^2 - 20x + 9 = 0 => x_1 ~~ -3/4, x_(2,3) ~~ 2/5 +- (1/5)i`
Since the derivative has the real root `x = -3/4` , you need to evaluate the value of the function at `f(-3/4) = f(-0.75)` , such that:
`f(-3/4) = 891/256 + 81/64 - 90/16 - 27/4 + 18 = 2655/256`
Evaluating the limit of the function at `+-oo` yields:
`lim_(x->+-oo)f(x) = +oo`
Hence, since there exists no change of sign of values of function over the real numbers set R, yields that the function `f(x) = 11x^4 - 3x^3 - 10x^2 + 9x + 18` has no zeroes for x in R.
Posted by sciencesolve on October 21, 2013 at 4:33 PM (Answer #1)
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