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.
Answer:
(y+1)/5
Step-by-step explanation:
you meant f^-1(x) ?
let y= 5x-1
Make x subject of formula
y+1=5x
x=(y+1)/5
f^-1(x)=(x+1)/5 replace y by x
Answer:
x = -3
Step-by-step explanation:
Answer:
-3
Step-by-step explanation:
30/(-2)-(-3)(4)
BIDMAS
-15 - (-3)(4)
-15 - -12
-15+12
-3
