Question

Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x3 − 6x2 − 15x + 2 (a) Find the interval on which f is increasing. (Enter your answer using interval notation.) Find the interval on which f is decreasing. (Enter your answer using interval notation.)

Answer 1

a)
`(-\infty, -1) \cup (5, \infty)`

b)
`(-1,5)`

Explanation:

The first step to solve this question is finding the roots of the derivative of x.

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

`ax^(2) + bx + c, a\\eq0`.

This polynomial has roots
`x_(1), x_(2)` such that
`ax^(2) + bx + c = a(x - x_(1))*(x - x_(2))`, given by the following formulas:

`x_(1) = (-b + √(\bigtriangleup))/(2*a)`

`x_(2) = (-b - √(\bigtriangleup))/(2*a)`

`\bigtriangleup = b^(2) - 4ac`

In this question:

`f(x) = x^(3) - 6x^(2) - 15x + 2`

So

`f'(x) = 3x^(2) - 12x - 15`

Finding the roots:

`3x^(2) - 12x - 15 = 0`

Simplifying by -3

`x^(2) - 4x - 5 = 0`

So
`a = 1, b = -4, c = -5`

Then

`\bigtriangleup = (-4)^(2) - 4*1*(-5) = 36`

`x_(1) = (-(-4) + √(36))/(2) = 5`

`x_(2) = (-(-4) - √(36))/(2) = -1`

So the function can be divided in three intervals.

They are:

Less than -1

Between -1 and 5

Higher than 5

In which it increases and which it decreases?

Less than -1

Lets find the derivative in a point in this interval, for example, -2

`f'(x) = 3x^(2) - 12x - 15`

`f'(-2) = 3*(-2)^(2) - 12*(-2) - 15 = 21`

Positive.

So in the interval of
`(-\infty, -1)`, the function increases.

Between -1 and 5

Will choose 0.

`f'(x) = 3x^(2) - 12x - 15`

`f'(0) = 3*(0)^(2) - 12*(0) - 15 = -15`

Negative.

So in the interval of
`(-1,5)`, the function decreases.

Higher than 5

Will choose 6.

`f'(x) = 3x^(2) - 12x - 15`

`f'(6) = 3*(6)^(2) - 12*(6) - 15 = 21`

Positive

So in the interval of
`(5, \infty)`, the function increases.

(a) Find the interval on which f is increasing.

Using interval notation

`(-\infty, -1) \cup (5, \infty)`

b) Find the interval on which f is decreasing.

`(-1,5)`