Respuesta :

calculista

we have

[tex]5y+4=(x+3)^{2}+\frac{1}{2}[/tex]

Step 1

Swap x and y

[tex]5x+4=(y+3)^{2}+\frac{1}{2}[/tex]

Step 2

Solve for y

[tex]5x+4-\frac{1}{2}=(y+3)^{2}[/tex]

[tex](y+3)^{2}=5x+\frac{7}{2}[/tex]

Square root both sides

[tex](y+3)=(+/-)\sqrt{5x+\frac{7}{2}}[/tex]

[tex]y=(+/-)\sqrt{5x+\frac{7}{2}}-3[/tex]

Step 3

Let

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

[tex]f(x)^{-1}=(+/-)\sqrt{5x+\frac{7}{2}}-3[/tex]

therefore

the answer is the option D

[tex]y=-3(+/-)\sqrt{5x+\frac{7}{2}}[/tex]

MrRoyal

The inverse of an equation is its opposite

The inverse equation is [tex]y = -3 \pm \sqrt{5x + \frac 72}[/tex]

The equation is given as:

[tex]5y + 4 = (x + 3)^2 + \frac12[/tex]

Subtract 1/2 from both sides

[tex]5y + \frac 72 = (x + 3)^2[/tex]

Take the square roots of both sides

[tex]\pm \sqrt{5y + \frac 72} = x+ 3[/tex]

Subtract 3 from both sides

[tex]-3 \pm \sqrt{5y + \frac 72} = x[/tex]

Rewrite the equation as:

[tex]x = -3 \pm \sqrt{5y + \frac 72}[/tex]

Swap x and y

[tex]y = -3 \pm \sqrt{5x + \frac 72}[/tex]

Hence, the inverse equation is [tex]y = -3 \pm \sqrt{5x + \frac 72}[/tex]

Read more about inverse equations at:

https://brainly.com/question/8120556

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