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gavmur [86]
3 years ago
8

SHOW YOUR WORK, PLEASE HELP!!!!!!!!!

Mathematics
1 answer:
Shkiper50 [21]3 years ago
3 0
To find the inverse, we swap the variables y and x, then solve for the new y.

3a. y=\frac{3}{x-1}

Swapping the variables: x=\frac{3}{y-1}
Solving for y: x(y-1)=3 \\ y-1= \frac{3}{x} \\ y=1+\frac{3}{x}
The domain of this inverse is x ≠ 0.
3b. y=x^2-1

Swapping: x = y^2 - 1
Solving for y: y^2 = x + 1 \\ y = \sqrt{x+1}
The domain of this inverse is x ≥ -1.
3c. y=\sqrt[3]{\frac{x-7}{3}}
Swapping: x=\sqrt[3]{\frac{y-7}{3}}
Solving for y: x^3=\frac{y-7}{3} \\ y-7=3x^3 \\ y=3x^3+7
The domain of this inverse is all real numbers.
4a. y=\frac{3}{x-1}, y=1+\frac{3}{x}
y=\frac{3}{(1+\frac{3}{x})-1} \\ y=\frac{3}{(\frac{3}{x})} \\ y=x
y=1+\frac{3}{(\frac{3}{x-1})} \\ y = 1+(x-1) \\ y = x

4c. y=\sqrt[3]{\frac{x-7}{3}}, y=3x^3+7
y=\sqrt[3]{\frac{(3x^3+7)-7}{3}} \\ y=\sqrt[3]{\frac{3x^3}{3}} \\ y=\sqrt[3]{x^3} \\ y=x
y=3(\sqrt[3]{\frac{x-7}{3}})^3+7 \\ y = 3({\frac{x-7}{3}})+7 \\ y = (x-7)+7 \\ y=x



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