QUESTION IMAGE
Question
which function is the inverse of $f(x)=x^3 - 6x^2 + 12x - 8$?
a. $f^{-1}(x)=sqrt3{x}-2$
b. $f^{-1}(x)=sqrt3{x}+2$
c. $f^{-1}(x)=sqrt3{x + 2}$
d. $f^{-1}(x)=\frac{sqrt3{sqrt{x^3 - 36x - 72}}}{2\times3^{\frac{1}{3}}}$
Step1: Simplify the original function
We start with the function \( f(x)=x^{3}-6x^{2}+12x - 8 \). Notice that this can be rewritten using the perfect cube formula \((a - b)^{3}=a^{3}-3a^{2}b + 3ab^{2}-b^{3}\). Let's check:
If we let \( a=x \) and \( b = 2 \), then \((x - 2)^{3}=x^{3}-3\times x^{2}\times2+3\times x\times2^{2}-2^{3}=x^{3}-6x^{2}+12x - 8\). So \( f(x)=(x - 2)^{3} \).
Step2: Find the inverse function
To find the inverse function, we first replace \( f(x) \) with \( y \), so \( y=(x - 2)^{3} \).
Then we swap \( x \) and \( y \) to get \( x=(y - 2)^{3} \).
Next, we solve for \( y \). Take the cube - root of both sides: \( \sqrt[3]{x}=y - 2 \).
Then, add 2 to both sides: \( y=\sqrt[3]{x}+2 \).
So the inverse function \( f^{-1}(x)=\sqrt[3]{x}+2 \).
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B. \( f^{-1}(x)=\sqrt[3]{x}+2 \)