is my answer correct?

Question

Find the intervals of increase and decrease for the function f(x) = x⁴ − 4x³ − 8x²− 1

f ' (x) = 4x³-12x²-16x

f'(-2) = -48

f'(-1) = 0

f'(0) = 0

f'(1) = -24

f'(2) = -48

f'(3) = -48

f'(4) = -48

f'(5) = 120

f'(6) = 336

x < 5

the value of y is negative which means it is decreasing

x > 5

the value of y is positive which mean it is increasing

this is my answer, is it correct?

Answer 1

Hey Girlie,

Good start with finding the derivative! What I would do next is find the critical points so that you can determine where the local minima and maxima are. This is helpful because the function will switch from increasing to decreasing at a local maximum, and from decreasing to increasing at a local minimum.

Critical points are values of x where the derivative of f(x) is zero (or undefined). So you can find the factors of the f'(x) polynomial to figure out what values of x satisfy this condition.

f'(x) = 4x³ - 12x² -16x

f'(x) = 4x(x² - 3x - 4)

f'(x) = 4x(x - 4)(x + 1)

Therefore, the values of x that make f'(x) equal to zero are 0, 4, and -1. These are the critical points of the original function. To figure out whether each critical point is a local minimum or maximum, you can find the second derivative at each point. At a local minimum, the curve is concave up, so the second derivative will have a positive value. At a max, it's concave down, so second derivative will be negative.

Let's calculate the second derivative at x = -1, x = 0, and x = 4.

f''(x) = 12x² - 24x - 16

f''(-1) = 20 --> positive, concave up, local minimum

f''(0) = -16 --> negative, concave down, local maximum

f''(4) = 80 --> positive, concave up, local minimum

Using this information, we can construct the intervals of increase and decrease for the original function. The boundaries of these intervals will be when the function switches from increasing to decreasing (or vice versa), which are the critical points. At a local minimum, a function switches from decreasing to increasing. At a local max, a function switches from increasing to decreasing. So the 3 critical points we identified are where these switches occur. Let's summarize (each sentence should technically begin with "immediately"):

To the left of x=-1, the function is decreasing. To the right of x=-1, the function is increasing.

To the left of x=0, the function is increasing. To the right of x=0, the function is decreasing.

To the left of x=4, the function is decreasing. To the right of x=4, the function is increasing.

So we can see that the function is first decreasing until -1, then increasing between -1 to 0, then decreasing from 0 to 4, then increasing after 4. Since this function's domain is all real numbers, we can extend the boundary trends to -∞ and ∞.

Final answer:

From -∞ to -1, the function is decreasing.

From -1 to 0, the function is increasing.

From 0 to 4, the function is decreasing.

From 4 to ∞, the function is increasing.

You can plot the equation online to confirm as well.

Let me know if this makes sense!

Best,

Kaleab

Answer 2

No, that's not how it works.

First locate the local extrema by setting the derivative equal to zero.

f ' (x) = 4x³-12x²-16x = 0

4x(x²-3x-4) = 0

4x(x - 4)(x + 1) = 0

x = -1, 0, 4

The following are the local extrema:

f(-1) = -4

f(0) = -1

f(4) = -129

If you recall your algebra about fourth degree polynomial with positive leading coefficient, 2 local minima and 1 local maximum, the graph should look like a letter "w". Therefore:

For the following interval of x:

x < -1 => decreasing

Point (-1,-4) => local minimum

-1< x < 0 => increasing

Point (0, -1) => local maximum

0 < x < 4 => decreasing

Point (4,-129) = local minimum

x > 4 => increasing

Answer 3

take the 1st derivative and set it = 0

f'(x)= 4x^3 -12x^2 -16x =0 factor

4x(x^2 -3x -4) =0

4x(x-4)(x+1) = 0

set each factor = 0 and solve for x

x=0, 4, and -1

those are the turning points

from -infinity to -1, the curve decreases

from -1 to 0, the curve increases

from 0 to 4 the curve decreases

from 4 to infinity the curve increases

the intervals alternate and you know that it goes to infinity, as x gets ever larger

so last interval is increasing. then alternate

Or use the 2nd derivative to see where it's increasing or decreasing

It's 4th degree, so there's 3 turning points which divide the intervals into 4 that alternate decreasing or increasing