Show that the equation 6x^5 + 13x + 1 = 0 has exactly one real root. (Use Rolle’s theorem and the intermediatd?
Show that the equation 6x^5 + 13x + 1 = 0 has exactly one
real root. (Use Rolle’s theorem and the intermediate- value
theorem.)
Update:where did you get the -1 and the 0 from? that's the part i don't understand
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Verified answer
f(x) = 6x^5 + 13x + 1
f'(x) = 30x^4 + 13 > 0
=> f(x) is increasing everywhere.
f(-1) = -6-13+1 = -18 < 0
f(0) = 1
So, by IVT there is exactly one real root for 6x^5 + 13x + 1 = 0.
Alternative method: By Descarte's rule of signs, the polynomial function has 0 positive real roots and 1 negative real root. Since 0 is not a root, there is exactly one real root.
Your method: f'(x) = 30x^4 + 13 = 0 has no real roots so the function is monotonic. f(-1) < 0 and f(1) > 0, so by the IVT, there is a zero on the interval (-1, 1). Since f is monotonic, this is the only root