9514 1404 393
Answer:
1) f⁻¹(x) = 6 ± 2√(x -1)
3) y = (x +4)² -2
5) y = (x -4)³ -4
Step-by-step explanation:
In general, swap x and y, then solve for y. Quadratics, as in the first problem, do not have an inverse function: the inverse relation is double-valued, unless the domain is restricted. Here, we're just going to consider these to be "solve for ..." problems, without too much concern for domain or range.
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1) x = f(y)
x = (1/4)(y -6)² +1
4(x -1) = (y-6)² . . . . . . subtract 1, multiply by 4
±2√(x -1) = y -6 . . . . square root
y = 6 ± 2√(x -1) . . . . inverse relation
f⁻¹(x) = 6 ± 2√(x -1) . . . . in functional form
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3) x = √(y +2) -4
x +4 = √(y +2) . . . . add 4
(x +4)² = y +2 . . . . square both sides
y = (x +4)² -2 . . . . . subtract 2
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5) x = ∛(y +4) +4
x -4 = ∛(y +4) . . . . . subtract 4
(x -4)³ = y +4 . . . . . cube both sides
y = (x -4)³ -4 . . . . . . subtract 4
Answer:
approx 58 yard we can determine it by pithagoras theorem
Answer:
$50.50 = 18 + 5n
Step-by-step explanation:
Answer:
11) Here given Function,
And, 
For f(x) = g(x)
When we solve this equation,
We found,
x = 12.5227 ≈ 12.53
Thus, the required solution is, x = 12.53
12) Here the height of rocket A in x second,
And, The height gain by the rocket B in x seconds,
If at x seconds both A and B gain the same height,
That is, f(x) = g(x)
⇒ 
⇒
⇒ 
⇒ 
⇒ x = 1.125 ≈ 1.13
Thus, the required solution is x = 1.13 seconds (approx)
Answer:
B
Step-by-step explanation:
