Please help me ASAP!

Range: 1 > x > 3

Step-by-step explanation:

We are given then function;

y = x³ - 6x² + 9x + 5

dy/dx = 3x² - 12x + 9

We want to find the range of values of x for which dy/dx > 0

Thus;

3x² - 12x + 9 > 0

Using quadratic formula, we have;

x = 1 and x = 3

Let's define the range;

try x = 0

3(0²) - 12(0) + 9 = 9

This is more than 0, so it is part of the solution.

Let's try x = 4

3(4²) - 12(4) + 9 = 9

This is more than 0 so it is part of the solution

Let's try x = 2

3(2²) - 12(2) + 9 = -3

This is less than 9 and so it's not part of the solution.

Thus the range of values of x for which dy/dx > 0 is;

1 > x > 3

x = 12

y = 15

step-by-step explanation: