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

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) + eˣ 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.

10/60

step-by-step explanation:

if you divide 10 by 60 the answer is 0.166666 while if you divide or solve the other problems it equals 0.6