Question

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).

Answer 1

Answer:

$f'(x) = 4x^2 - 3$ for $x \le -3$

Step-by-step explanation:

See attachment for proper question

Given

$f(x) = -\frac{1}{2}\sqrt{x + 3}$

For

$x \ge -3$

Required

Determine the inverse function

$f(x) = -\frac{1}{2}\sqrt{x + 3}$

Replace f(x) with y

$y = -\frac{1}{2}\sqrt{x + 3}$

Swap the positions of x and y

$x = -\frac{1}{2}\sqrt{y + 3}$

Multiply both sides by -2

$-2 * x =-2 * -\frac{1}{2}\sqrt{y + 3}$

$-2x =\sqrt{y + 3}$

Square both sides

$(-2x)^2 =(\sqrt{y + 3})^2$

$4x^2 =y + 3$

Make y the subject

$y = 4x^2 - 3$

The inverse has been solved. So, we need to replace y with f'(x)

$f'(x) = 4x^2 - 3$

Next, is to determine the interval

$x \ge -3$

Change inequality to $\le$

$x \le -3$

Hence, the inverse function is:

$f'(x) = 4x^2 - 3$ for $x \le -3$