Answer:
[tex]h^{-1}(x)=x^3+6x^2+12x+7[/tex]
Explanation:
Given the below function;
[tex]h(x)=\sqrt[3]{x+1}-2[/tex]
We'll follow the below steps to determine the inverse of the above function;
Step 1: Replace h(x) with y;
[tex]y=\sqrt[3]{x+1}-2[/tex]
Step 2: Switch x and y;
[tex]x=\sqrt[3]{y+1}-2[/tex]
Step 3: Solve for y by first adding 2 to both sides;
[tex]\begin{gathered} x+2=\sqrt[3]{y+1}-2+2 \\ x+2=\sqrt[3]{y+1} \end{gathered}[/tex]
Step 4: Take the cube of both sides;
[tex]\begin{gathered} (x+2)^3=(\sqrt[3]{y+1})^3 \\ (x+2)^3=y+1 \end{gathered}[/tex]
Step 5: Expand the cube power;
Recall;
[tex](a+b)^3=a^3+3a^2b+3ab^2+b^3[/tex]
Applying the above, we'll have;
[tex]\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}[/tex]
Step 6: Subtract 1 from both sides of the equation;
[tex]\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}[/tex]
Step 7: Replace y with h^-1(x);
[tex]h^{-1}(x)=x^3+6x^2+12x+7[/tex]
