Answer:
[tex]f'(x) = 4x^2 - 3[/tex] for [tex]x \le -3[/tex]
Step-by-step explanation:
See attachment for proper question
Given
[tex]f(x) = -\frac{1}{2}\sqrt{x + 3}[/tex]
For
[tex]x \ge -3[/tex]
Required
Determine the inverse function
[tex]f(x) = -\frac{1}{2}\sqrt{x + 3}[/tex]
Replace f(x) with y
[tex]y = -\frac{1}{2}\sqrt{x + 3}[/tex]
Swap the positions of x and y
[tex]x = -\frac{1}{2}\sqrt{y + 3}[/tex]
Multiply both sides by -2
[tex]-2 * x =-2 * -\frac{1}{2}\sqrt{y + 3}[/tex]
[tex]-2x =\sqrt{y + 3}[/tex]
Square both sides
[tex](-2x)^2 =(\sqrt{y + 3})^2[/tex]
[tex]4x^2 =y + 3[/tex]
Make y the subject
[tex]y = 4x^2 - 3[/tex]
The inverse has been solved. So, we need to replace y with f'(x)
[tex]f'(x) = 4x^2 - 3[/tex]
Next, is to determine the interval
[tex]x \ge -3[/tex]
Change inequality to [tex]\le[/tex]
[tex]x \le -3[/tex]
Hence, the inverse function is:
[tex]f'(x) = 4x^2 - 3[/tex] for [tex]x \le -3[/tex]
