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23 paź 2020 · #MA8151#engineeringmathematics MA8151 ENGINEERING MATHEMATICS – I https://alexmathsonlineeducation.blogspot.com/p/engineering-mathematics-i.html https://alex...
9 maj 2018 · Explanation: This is equivalent to showing that the function f (x) = x3 − 15x + c cannot have two zeros in [ −2,2]. Summary: If f had two zeros in [ − 2,2], then 3x2 − 15 = 0 would have a solution in [ −2,2]. But 3x2 −15 = 0 does not have any such solution. Therefore, there f cannot have two different zeros in [ − 2,2]. Details:
7 maj 2016 · Assume there are $2$ or more real roots, then by the Rolle's theorem: $f'(c) = 0$ in $[-2,2]$, but $f'(c) = 3c^2 - 15 = 3(c^2-5) < 0$ since $|c| \leq 2$. Thus it can't happen.
22 sty 2015 · Step 1: The equation is and the interval . Consider . The function is is continuous on the closed interval . Consider then , It follows that and . Therefore apply the intermediate theorem to state that there must be some k in Such that f (k)=c . has atmost one zero in the closed interval . Solution: has atmost one zero in the closed interval .
This tells you that if f(x) = x3 − 15x + c and f has two zeros somewhere (zeroes of f are exactly the roots of your equation) then its derivative has a zero in between. Now, in this case, f′(x) = 3x2 − 15, which is zero at − 5–√ and 5–√. In particular, f′ is never zero in the interval [−2, 2].
To show that the equation x^3-15x+c=0 has at most one root in the interval (-2, 2), we can use the concept of the Intermediate Value Theorem. We need to examine the behavior of the function f(x) = x^3 - 15x + c as x approaches -2 and 2 from either side of the interval.
15 mar 2023 · Roots of a cubic Show that the equation $x^{3}-15 x+1=0$ has three solutions in the interval $[-4,4]$ . Video Solution, solved step-by-step from our expert human educators: Show that the equation $x^{3}-15 x+c=0$ has at most one root in the interval
