Please I need like an explanation of how it was done.
2 answers:
Answer:
the point is (3, 3) . . . . . . y = 3
Step-by-step explanation:
Write the point-slope equation of the line through the point you know. Then evaluate that equation for x=3 to see what the value of y is.
Point-slope form:
y = m(x -h) +k . . . . slope m through point (h, k)
y = -1/2(x -9) +0 . . . . line with slope -1/2 through point (9, 0)
For x=3, the value of y is ...
y = -1/2(3 -9) + 0 = -1/2(-6) = 3
The value of y is 3.
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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, ∞)