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Flura [38]
3 years ago
14

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

Mathematics
1 answer:
ExtremeBDS [4]3 years ago
5 0

If <em>x</em> = -1, you have

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

and if <em>x</em> = 0, you have

2(0) + 3 cos(0) + <em>e</em> ⁰ = 4 > 0

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

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