For y=(1/4)x^4-(2/3)x^3+(1/2)x^2-3, find the exact intervals on which the function is

a. increasing
b. decreasing
c. concave up
d. concave down
Then find any
e. local extreme values
f. inflection points

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I thought it'd be better to find extreme values first, then you can find the points where the function is increasing or decreasing. (I'll worry about concavity and inflection points later, since that's second derivative)

So I found y':
x^3-2x^2-x
...and set it equal to zero to find the points where derivative crosses the x-axis
x^3-2x^2-x=0
I factored and used the quadratic formula, and got (2+sqrt8)/2 and (2-sqrt8)/2, or -.414 and 2.414. Oh, and zero too because I factored out an x.

So my problem is that the answers (which were given to me ahead of time to check my work) don't match my work. Apparently y is increasing from [0, infinity) and decreasing from (-infinity, 0], and none of the points I found from finding y'.

Can you find what I did wrong?

Thanks!

Your derivative is wrong,

you should have
y' = x^3 - 2x^2 + x

When I set x^3 - 2x^2 - x = 0, I get
x(x^2 - 2x + 1 ) = 0
x(x-1)(x-1) = 0

x = 0 or x = 1

take it from there.

Oh wow, such a simple mistake throws off everything, haha. Thanks

To find the intervals on which the function is increasing or decreasing, we need to analyze the sign of the derivative, which is denoted by y'.

First, let's find the derivative correctly:

Given: y = (1/4)x^4 - (2/3)x^3 + (1/2)x^2 - 3

To find y', we differentiate each term with respect to x:

y' = d/dx[(1/4)x^4 - (2/3)x^3 + (1/2)x^2 - 3]
= (1/4)(4x^3) - (2/3)(3x^2) + (1/2)(2x^1)
= x^3 - 2x^2 + x

Now let's solve for y' = 0 to find the critical points (where the derivative crosses the x-axis):

x^3 - 2x^2 + x = 0

Factoring out an x, we get:

x(x^2 - 2x + 1) = 0

The quadratic expression can be factored further:

x(x - 1)(x - 1) = 0

So we have three critical points: x = 0 and x = 1 (with multiplicity 2).

Now, let's analyze the sign of the derivative in the intervals between and around the critical points.

1. For x < 0:
Substitute a test value, e.g., -1, into y':
y'(-1) = (-1)^3 - 2(-1)^2 + (-1) = -1 - 2 + (-1) = -4
Since y'(-1) is negative, the function is decreasing in this interval.

2. For 0 < x < 1:
Substitute a test value, e.g., 0.5, into y':
y'(0.5) = (0.5)^3 - 2(0.5)^2 + (0.5) = 0.125 - 0.5 + 0.5 = 0.125
Since y'(0.5) is positive, the function is increasing in this interval.

3. For x > 1:
Substitute a test value, e.g., 2, into y':
y'(2) = (2)^3 - 2(2)^2 + (2) = 8 - 8 + 2 = 2
Since y'(2) is positive, the function is increasing in this interval.

Therefore, we have determined the intervals on which the function y = (1/4)x^4 - (2/3)x^3 + (1/2)x^2 - 3 is increasing and decreasing:

a. Increasing: (0, infinity)
b. Decreasing: (-infinity, 0)

Now let's move on to finding the concave up and concave down intervals and the local extreme values.

To find concavity and inflection points, we need to analyze the second derivative of the function (the derivative of the derivative).

To find y'', we differentiate y':

y'' = d/dx[x^3 - 2x^2 + x]
= 3x^2 - 4x + 1

Now let's set y'' = 0 to find the points where the second derivative crosses the x-axis:

3x^2 - 4x + 1 = 0

This quadratic equation can be factored as:

(3x - 1)(x - 1) = 0

So we have two critical points for the second derivative: x = 1 and x = 1/3.

Now we can determine the intervals of concavity by analyzing the sign of the second derivative in the intervals between and around the critical points.

1. For x < 1/3:
Substitute a test value, e.g., 0, into y'':
y''(0) = 3(0)^2 - 4(0) + 1 = 1
Since y''(0) is positive, the function is concave up in this interval.

2. For 1/3 < x < 1:
Substitute a test value, e.g., 0.5, into y'':
y''(0.5) = 3(0.5)^2 - 4(0.5) + 1 = 0.75
Since y''(0.5) is positive, the function is concave up in this interval.

3. For x > 1:
Substitute a test value, e.g., 2, into y'':
y''(2) = 3(2)^2 - 4(2) + 1 = 5
Since y''(2) is positive, the function is concave up in this interval.

Therefore, we have determined the intervals of concavity:

c. Concave up: (-infinity, 1/3) U (1, infinity)
d. Concave down: (1/3, 1)

To find the local extreme values (points where the function has a maximum or minimum), we need to examine the behavior of the function around the critical points and within the given intervals.

The critical points are:
- x = 0 (found from y' = 0)
- x = 1 (found from y' = 0, with multiplicity 2)

To determine if these points are local maximum or minimum, we can use the first or second derivative test.

For x = 0:
Substitute x = 0 into y:
y(0) = (1/4)(0)^4 - (2/3)(0)^3 + (1/2)(0)^2 - 3 = -3
Since there is no interval around x = 0 where the function changes from increasing to decreasing or vice versa, there is no local extreme value at x = 0.

For x = 1:
Substitute x = 1 into y:
y(1) = (1/4)(1)^4 - (2/3)(1)^3 + (1/2)(1)^2 - 3 = -9/12 = -3/4
Since there is a change in behavior from increasing to decreasing at x = 1, there is a local maximum at (1, -3/4).

Therefore, we have determined the local extreme values:

e. Local extreme values: (1, -3/4) - local maximum

Finally, to find the inflection points (the points where the concavity changes), we need to look for the points where the second derivative changes sign.

The critical point for the second derivative is:
- x = 1/3 (found from y'' = 0)

Substitute x = 1/3 into y:
y(1/3) = (1/4)(1/3)^4 - (2/3)(1/3)^3 + (1/2)(1/3)^2 - 3 = -499/216

Therefore, we have determined the inflection point:

f. Inflection point: (1/3, -499/216)

In summary, the exact intervals for the function y = (1/4)x^4 - (2/3)x^3 + (1/2)x^2 - 3 are:

a. Increasing: (0, infinity)
b. Decreasing: (-infinity, 0)
c. Concave up: (-infinity, 1/3) U (1, infinity)
d. Concave down: (1/3, 1)

With the local extreme value at (1, -3/4) as a local maximum and the inflection point at (1/3, -499/216).