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3 years ago

12

30 POINTS IF SOMEONE CAN SOLVE THIS ALGEBRA II PROBLEM

1 answer:

Oksana_A [137]3 years ago

7 0

V= 1.34 sqrt (Ln- 10)
Ln > 10

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Y_Kistochka [10]

The function $f(x)=(x-2)^3+8$ is one-to-one and the inverse of $f(x)$ is $f^{-1} (x)=(x-8)^\frac{1}{3}+2$ .

A function is one-to-one if for $f(a)=f(b)$ , we show that $a=b$ .

$f(a)=(a-2)^3+8$

$f(b)=(b-2)^3+8$

Take $f(a)=f(b)$

$(a-2)^3+8=(b-2)^3+8$

$(a-2)^3=(b-2)^3$

Let $A=a-2$ and $B=b-2$

If $A^3=B^3$ , then $A=B$ .

So, $a-2=b-2$

$a=b$ .

So, the function is one-to-one.

For the inverse of $f(x)$ , replace $x$ with $y$ and solve for $y$ .

$x=(y-2)^3+8$

$(y-2)^3=x-8$

$y-2=(x-8)^\frac{1}{3}$

$y=(x-8)^\frac{1}{3}+2$

So, the inverse of $f(x)$ is $y=(x-8)^\frac{1}{3}+2$ .

Learn more about and inverse of a function here:

brainly.com/question/2541698?referrer=searchResults

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2 years ago