Question #145760
Using the Intermediate Value Theorem and a calculator, find an interval of length 0.01 that contains a root of x5−x2+2x+3=0, rounding off interval endpoints to the nearest hundredth.

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Expert's answer
2020-12-01T01:28:59-0500
SolutionSolution

For f(x)=x5x2+2x+3f(x) = x^5 - x^2 + 2x + 3;


f(0)=3f(1)=112+3=1f(0) = 3\\ f(-1) = -1 - 1 - 2 + 3 = -1


Thus, as this is a continuous function, from the Intermediate Value Theorem, we know that, as f(1)<0f(-1) < 0 and f(0)>0f(0) > 0, there is a root between 1-1 and 0.0.


While we could use the bisection method, the easiest thing is to find f(x)f(x) for xx between 1-1 and in increments of .01.01


f(0.88)=0.0621319168000003f(0.87)=0.00467907929999978f(-0.88) = -0.0621319168000003\\ f(-0.87) = 0.00467907929999978


Then, the Intermediate Value Theorem tells us, at f(0.88)<0f(-0.88) < 0 and f(0.87)=0.00467907929999978f(-0.87) = 0.00467907929999978, there is a root in the interval (0.88,0.87)(-0.88, -0.87)


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