Answer:
[tex]f^{-1}(x)=-3(+/-)\sqrt{\frac{10x+7}{2}}[/tex]
Step-by-step explanation:
we have
[tex]5y+4=(x+3)^{2}+\frac{1}{2}[/tex]
Exchange x for y and y for x
[tex]5x+4=(y+3)^{2}+\frac{1}{2}[/tex]
Isolate the variable y
[tex]5x+4-\frac{1}{2}=(y+3)^{2}[/tex]
[tex]5x+\frac{7}{2}=(y+3)^{2}[/tex]
[tex]\frac{10x+7}{2}=(y+3)^{2}[/tex]
Take square root both sides
[tex](y+3)=(+/-)\sqrt{\frac{10x+7}{2}}[/tex]
[tex]y=-3(+/-)\sqrt{\frac{10x+7}{2}}[/tex]
Let
[tex]f^{-1}(x)=y[/tex]
[tex]f^{-1}(x)=-3(+/-)\sqrt{\frac{10x+7}{2}}[/tex]
