Question

Use the IVT to prove that one solution to
`x^(3) +x-5=0` is between x=1 and x=2.

Please show work and if your answer makes no sense I will report. Thanks

Answer 1

Let f(x) = x³ + x - 5. f(x) is a polynomial so it's continuous everywhere on its domain (all real numbers). Since

f (1) = 1³ + 1 - 5 = -3 < 0

and

f (2) = 2³ + 2 - 5 = 5 > 0

it follows by the intermediate value theorem that there at least one number x = c between 1 and 2 for which f(c) = 0.