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Let f(x) = (3x + 9. Find f-'(x).=Vf-'(x) =help (formulas)

Let Fx 3x 9 Find FxVfx Help Formulas class=

Sagot :

Given the function:

[tex]f\mleft(x\mright)=\sqrt{3x+9}[/tex]

You can find the Inverse Function by following the steps shown below:

1. Rewrite the function using:

[tex]f(x)=y[/tex]

Then:

[tex]y=\sqrt{3x+9}[/tex]

2. Solve for "x":

- Square both sides of the equation, in order to undo the effect of the square root on the right side:

[tex]\begin{gathered} (y)^2=(\sqrt[]{3x+9})^2 \\ y^2=3x+9 \end{gathered}[/tex]

- Apply the Subtraction Property of Equality by subtraction 9 from both sides of the equation:

[tex]\begin{gathered} y^2-(9)=3x+9-(9) \\ \\ y^2-9=3x \end{gathered}[/tex]

- Apply the Division Property of Equality by dividing both sides of the equation by 3:

[tex]\begin{gathered} \frac{y^2-9}{3}=\frac{3x}{3} \\ \\ \frac{y^2-9}{3}=x \\ \\ x=\frac{y^2-9}{3} \end{gathered}[/tex]

3. Swap the variables:

[tex]y=\frac{x^2-9}{3}[/tex]

4. Replace the variable "y" with:

[tex]y=f^{-1}(x)[/tex]

Then, you get:

[tex]f^{-1}(x)=\frac{x^2-9}{3}[/tex]

Hence, the answer is:

[tex]f^{-1}(x)=\frac{x^2-9}{3}[/tex]