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.

x=17

step-by-step explanation:

solve 3x-24=x+10

then that makes it 2x=34 because you subtract an x from both sides to get rid of the one on the right, and then you add 24 to both sides to get rid of the 24 on the left

divide by both sides by 2 to get your answer so that you can get your answer

step-by-step explanation:

bob the banana