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

9514 1404 393

Answer:

  f^-1(x) = ∛((x +5)/2)

Step-by-step explanation:

Continue to solve for y.

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

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

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

Answer 2

• 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

$\\ \rm\Rrightarrow y^3=\dfrac{x+5}{2}$

$\\ \rm\Rrightarrow y=\sqrt[3]{\dfrac{x+5}{2}}$

• This is the inverse

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