Answer: c.(3, 25) and (7, 9)
y = –x^2 + 6x + 16 and y = –4x + 37
Plug in -4x+37 for y in first equation . It becomes
Combine like terms. add 4x and subtract 37 on both sides
Divide the whole equation by -1 to remove negative sign from -x^2
Now factor the left hand side
(x-7)(x-3) = 0
x-7 =0 and x-3=0
x= 7 and x=3
Now we find out y using y = –4x + 37
when x= 7 , then y=-4(7) +37 = 9
when x= 3, then y=-4(3) + 37 = 25
We write solution set as (x,y)
(7,9) and (3,25) is our solution set
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, ∞)