Question

Can you help me to do this please?Determine the inverse of the h(x)

Answer 1

`h^(-1)(x)=x^3+6x^2+12x+7`

Step-by-step explanation:

Given the below function;

`h(x)=\sqrt[3]{x+1}-2`

We'll follow the below steps to determine the inverse of the above function;

Step 1: Replace h(x) with y;

`y=\sqrt[3]{x+1}-2`

Step 2: Switch x and y;

`x=\sqrt[3]{y+1}-2`

Step 3: Solve for y by first adding 2 to both sides;

`\begin{gathered} x+2=\sqrt[3]{y+1}-2+2 \\ x+2=\sqrt[3]{y+1} \end{gathered}`

Step 4: Take the cube of both sides;

`\begin{gathered} (x+2)^3=(\sqrt[3]{y+1})^3 \\ (x+2)^3=y+1 \end{gathered}`

Step 5: Expand the cube power;

Recall;

`(a+b)^3=a^3+3a^2b+3ab^2+b^3`

Applying the above, we'll have;

`\begin{gathered} (x+2)^3=y+1 \\ x^3+3x^2\cdot2+3x\cdot2^2+2^3=y+1 \\ x^3+6x^2+12x+8=y+1 \end{gathered}`

Step 6: Subtract 1 from both sides of the equation;

`\begin{gathered} x^3+6x^2+12x+8-1=y+1-1 \\ x^3+6x^2+12x+7=y \\ \therefore y=x^3+6x^2+12x+7 \end{gathered}`

Step 7: Replace y with h^-1(x);

`h^(-1)(x)=x^3+6x^2+12x+7`