SOLUTION: The function f defined on (−∞,∞) by
f(x)=22−58x−14(8+29x)7
is one-to-one and has range (−∞,∞).
Find f−1(38).
Answer :
(Do NOT USE DECIMALS if your answ
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-> SOLUTION: The function f defined on (−∞,∞) by
f(x)=22−58x−14(8+29x)7
is one-to-one and has range (−∞,∞).
Find f−1(38).
Answer :
(Do NOT USE DECIMALS if your answ
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Question 1170516: The function f defined on (−∞,∞) by
f(x)=22−58x−14(8+29x)7
is one-to-one and has range (−∞,∞).
Find f−1(38).
Answer :
(Do NOT USE DECIMALS if your answer is a fraction or an irrational number) Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let $f(x) = 22 - 58x - 14(8+29x)^7$.
We want to find $f^{-1}(38)$.
Let $y = f(x)$.
We want to find $x$ such that $f(x) = 38$.
So we need to solve the equation:
$38 = 22 - 58x - 14(8+29x)^7$.
$38 - 22 = -58x - 14(8+29x)^7$.
$16 = -58x - 14(8+29x)^7$.
$16 + 58x = -14(8+29x)^7$.
$-8 - 29x = 7(8+29x)^7$.
Let $u = 8+29x$. Then $-u = 7u^7$.
$u^7 = -\frac{u}{7}$.
$u^7 + \frac{u}{7} = 0$.
$u(u^6 + \frac{1}{7}) = 0$.
Since $u^6 + \frac{1}{7} > 0$ for all real $u$, we must have $u = 0$.
$8+29x = 0$.
$29x = -8$.
$x = -\frac{8}{29}$.
Now, we verify this value.
$f(-\frac{8}{29}) = 22 - 58(-\frac{8}{29}) - 14(8+29(-\frac{8}{29}))^7$.
$= 22 + 2(8) - 14(8-8)^7$.
$= 22 + 16 - 14(0)^7$.
$= 38 - 0$.
$= 38$.
Thus, $f^{-1}(38) = -\frac{8}{29}$.
Final Answer: The final answer is $\boxed{-8/29}$
