Let <em>f(x)</em> = <em>x</em>³ + <em>x</em> - 5. <em>f(x)</em> is a polynomial so it's continuous everywhere on its domain (all real numbers). Since
<em>f</em> (1) = 1³ + 1 - 5 = -3 < 0
and
<em>f</em> (2) = 2³ + 2 - 5 = 5 > 0
it follows by the intermediate value theorem that there at least one number <em>x</em> = <em>c</em> between 1 and 2 for which <em>f(c)</em> = 0.
I hope this helps you
(f (x))^2= (square root of x-2)^2
f^2 (x)=x-2
x=f^2 (x)+2
f^-1 (x)=x^2+2
Answer:
Step-by-step explanation:
A) 950, 900, 1000
b) 1860, 1900, 2000
c) 200, 200, 0
d) 650, 600, 1000
e) 19900, 19900, 20000
