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Show that the equation 2x + 3 cos x + e ^ x = 0 has a root on the interval [- 1, 0]

User Jveazey
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1 Answer

21 votes
21 votes

If x = -1, you have

2(-1) + 3 cos(-1) + e ⁻¹ ≈ -0.0112136 < 0

and if x = 0, you have

2(0) + 3 cos(0) + e ⁰ = 4 > 0

The function f(x) = 2x + 3 cos(x) + is continuous over the real numbers, so the intermediate value theorem applies, and it says that there is some -1 < c < 0 such that f(c) = 0.

User Optimworks
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