Note that f(x) as given is <em>not</em> invertible. By definition of inverse function,


which is a cubic polynomial in
with three distinct roots, so we could have three possible inverses, each valid over a subset of the domain of f(x).
Choose one of these inverses by restricting the domain of f(x) accordingly. Since a polynomial is monotonic between its extrema, we can determine where f(x) has its critical/turning points, then split the real line at these points.
f'(x) = 3x² - 1 = 0 ⇒ x = ±1/√3
So, we have three subsets over which f(x) can be considered invertible.
• (-∞, -1/√3)
• (-1/√3, 1/√3)
• (1/√3, ∞)
By the inverse function theorem,

where f(a) = b.
Solve f(x) = 2 for x :
x³ - x + 2 = 2
x³ - x = 0
x (x² - 1) = 0
x (x - 1) (x + 1) = 0
x = 0 or x = 1 or x = -1
Then
can be one of
• 1/f'(-1) = 1/2, if we restrict to (-∞, -1/√3);
• 1/f'(0) = -1, if we restrict to (-1/√3, 1/√3); or
• 1/f'(1) = 1/2, if we restrict to (1/√3, ∞)
Answer:
n = 8 or n = −6
Explanation:
First, isolate the absolute value:
5 − |n − 1| = −2
7 − |n − 1| = 0
7 = |n − 1|
Then, split this into two equations, since the result of the absolute value could be positive or negative:
7 = n − 1 7 = −(n − 1)
8 = n 7 = −n + 1
n = −6
use three cups of flour that is 11% whole wheat and 9 cups of 63%
Answer:
(I cant put the line in the middle)
7
72
Step-by-step explanation: