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:
she is incorrect because if she was given a 20% discount the amount she would have to pay is 38.4
Step-by-step explanation:
48 x .2 = 9.6
48 - 9.6 = 38.4 not 40
Answer:
-1
Step-by-step explanation:
1-2k=-3k
1-2k+2k=-3k+2k
1=-1k
1/-1=-1k/-1
-1=k
Answer:x>3 1/6
Step-by-step explanation:
5/12-(x-3)/6 <(x-2)/3
5/12 *12-(x-3)/6*12<(x-2)/6*12
5-2(x-3)<4(x-2)
5-2x+6<4x-8
-2x+11<4x-8
-2x<4x-19
-6x<-19
-6x(-1)<-19(-1)
6x>19
x>19/6
x>3 1/6
just change all the inequality signs to a x is greater or equal to 3 1/6 and on the rest change to and < and equal sign only put > or equal sign on the last three lines
Answer:
b/(b+a)
Step-by-step explanation:
(1/a)-(1/b) :[ (b²-a²)/ab²]
first solve :
common denominator ab
(1/a)-(1/b) = (b-a)/ab
[b-a/ab] : [(b²-a²)/ab²]
when divide fraction ( division sign turn to (×) and flip the second fraction(reciprocal):
[b-a/ab] × [ab²/ (b²-a²)]
then simplify : ab²/ab = b
(b-a)×(b/b²-a²)
factorize : b²-a² = (b-a)(b+a)
(b-a)×(b/(b-a)(b+a)) simplify : (b-a)/b-a = 1
[(b-a)(b)]/[(b-a)(b+a)
b/b+a
