Question

The graph of F ( x ) = 4/x2 - 2x - 3 is shown
for which values of X is f ( x ) decreasing

Answer 1

Hi there!

$\large\boxed{(-\infty, \sqrt[3]{-4}) \text{ and } (0, \infty) }$

We can find the values of x for which f(x) is decreasing by finding the derivative of f(x):

$f(x) = \frac{4}{x^2} - 2x - 3$

Taking the derivative gets:

$f'(x) = -\frac{8}{x^3} - 2$

Find the values for which f'(x) < 0 (less than 0, so f(x) decreasing):

0 = -8/x³ - 2

2 = -8/x³

2x³ = -8

x³ = -4

$x = \sqrt[3]{-4}$

Another critical point is also where the graph has an asymptote (undefined), so at x = 0.

Plug in points into the equation for f'(x) on both sides of each x value to find the intervals for which the graph is less than 0:

f'(1) = -8/1 - 2 = -10 < 0

f'(-1) = -8/(-1) - 2 = 6 > 0

f'(-2) = -8/-8 - 2 = -1 < 0

Thus, the values of x are:

$(-\infty, \sqrt[3]{-4}) \text{ and } (0, \infty)$