Question

Find the inverse of the function f(x) = 2x^3 - 5

This is what I have so far
F(x) = 2x^3 - 5
Y = 2x^3 - 5
X = 2y^3 - 5
+5 +5
x + 5 = 2y^3

Answer 1

• y be f(x)

`\\ \rm\Rrightarrow y=2x^3-5`

• Interchange y and x

`\\ \rm\Rrightarrow x=2y^3-5`

`\\ \rm\Rrightarrow 2y^3=x+5`

• Isolate y

`\\ \rm\Rrightarrow y^3=(x+5)/(2)`

`\\ \rm\Rrightarrow y=\sqrt[3]{(x+5)/(2)}`

• This is the inverse

`\\ \rm\Rrightarrow f^(-1)(x)=\sqrt[3]{(x+5)/(2)}`

Answer 2

9514 1404 393

Answer:

f^-1(x) = ∛((x +5)/2)

Explanation:

Continue to solve for y.

`x +5=2y^3\qquad\text{so far so good}\\\\(x+5)/(2)=y^3\qquad\text{divide by 2}\\\\\sqrt[3]{(x+5)/(2)}=y\qquad\text{take the cube root}\\\\\boxed{f^(-1)(x)=\sqrt[3]{(x+5)/(2)}}\qquad\text{expressed in functional form}`

If you want to "rationalize the denominator", you can multiply both numerator and denominator by 4 inside the radical. This will simplify to ...

`\displaystyle\boxed{f^(-1)(x)=(1)/(2)\sqrt[3]{4x+20}}`