Answer:
1) [tex]\therefore f^{-1}(x)=\frac{1}{2}x^2-3,\:x\ge0[/tex].
2) [tex]f^{-1}(x)=\sqrt[4]{x-16},\:x\ge16[/tex]
3) [tex]f^{-1}(x)=\frac{8}{3}x-\frac{8}{3}[/tex]
Step-by-step explanation:
1) The given function is [tex]f(x)=\sqrt{2x=6},\:\:x\ge-3[/tex].
Let [tex]y=\sqrt{2x-6}[/tex].
Interchange x and y to get:
[tex]x=\sqrt{2y+6}[/tex].
Square both sides;
[tex]x^2=2y+6[/tex].
Solve for y.
[tex]x^2-6=2y[/tex].
[tex]\frac{1}{2}x^2-3=y[/tex].
[tex]\therefore f^{-1}(x)=\frac{1}{2}x^2-3,\:\:x\ge0[/tex].
2) The given function is [tex]f(x)=x^4+16,\:\:x\ge0[/tex]
Let [tex]y=x^4+16[/tex]
Interchange x and y;
[tex]x=y^4+16[/tex]
Solve for y
[tex]x-16=y^4[/tex]
[tex]\sqrt[4]{x-16}=y[/tex],
Therefore the inverse is
[tex]f^{-1}(x)=\sqrt[4]{x-16},\:x\ge16[/tex]
3) The given function is [tex]f(x)=\frac{3x}{8}+1[/tex]
Let [tex]y=\frac{3x}{8}+1[/tex]
Interchange x and y
[tex]x=\frac{3y}{8}+1[/tex]
Solve for y:
[tex]x-1=\frac{3y}{8}[/tex]
[tex]8x-8=3y[/tex]
[tex]\frac{8}{3}x-\frac{8}{3}=y[/tex]
Therefore the inverse is
[tex]f^{-1}(x)=\frac{8}{3}x-\frac{8}{3}[/tex]
